IPMAT Logarithms PYQs

Q1:ipmat indore 2024QA › LogarithmsMediumSA · TITA

If

4^{\log_2{x}} - 4x + 9^{\log_3{y}} - 16y + 68 = 0

, then

y - x

equals:

Enter your answer to attempt

The Setup: We are given the equation

4^{\log_{2}x}-4x+9^{\log_{3}y}-16y+68=0

. We need to evaluate the expression

y-x

. Step 1: Simplify the logarithmic terms using the base-change exponent identity

a^{\log_a b} = b

. For the first term:

4^{\log_{2}x} = (2^2)^{\log_{2}x} = 2^{2\log_{2}x} = 2^{\log_{2}(x^2)} = x^2

For the third term:

9^{\log_{3}y} = (3^2)^{\log_{3}y} = 3^{2\log_{3}y} = 3^{\log_{3}(y^2)} = y^2

Step 2: Substitute the simplified terms back into the algebraic equation.

x^2 - 4x + y^2 - 16y + 68 = 0

Step 3: Complete the square for both the

x

and

y

variables. Isolate the respective variables and add the required constants:

(x^2 - 4x + 4) + (y^2 - 16y + 64) = 0

(x - 2)^2 + (y - 8)^2 = 0

Note that

4 + 64 = 68

, which perfectly balances the original constant. Step 4: Solve for

x

and

y

. The sum of two real squares equals zero if and only if each independent square evaluates to zero.

x - 2 = 0 \Rightarrow x = 2

y - 8 = 0 \Rightarrow y = 8

Both values are strictly positive, satisfying the logarithmic domain restrictions. Step 5: Calculate the final target expression

y - x

.

8 - 2 = 6

Final Answer: 6

Q2:ipmat indore 2022QA › LogarithmsMediumSA · TITA

If

\log _{\left(x^{2}\right)} y+\log _{\left(y^{2}\right)} x=1

and

y=x^{2}-30

, then the value of

x^{2}+y^{2}

is ___________.

Enter your answer to attempt

The Setup: We are given a logarithmic equation involving bases of

x^2

and

y^2

, alongside a quadratic relation

y=x^2-30

. We must simplify the log equation to find

x

and

y

, then compute

x^2+y^2

. Step 1: Simplify the logarithmic equation. Use the base power rule

\log_{(b^k)} a = \frac{1}{k} \log_b a

:

\frac{1}{2}\log_x y + \frac{1}{2}\log_y x = 1

Multiply by 2:

\log_x y + \log_y x = 2

Step 2: Solve for the relationship between

x

and

y

. Let

a = \log_x y

. By logarithmic inversion,

\log_y x = \frac{1}{a}

.

a + \frac{1}{a} = 2

a^2 - 2a + 1 = 0 \implies (a - 1)^2 = 0 \implies a = 1

Therefore,

\log_x y = 1

, which strictly implies

y = x

. Step 3: Apply the quadratic relation to find the valid coordinates. Substitute

y = x

into the second equation

y = x^2 - 30

:

x = x^2 - 30 \implies x^2 - x - 30 = 0

Factor the quadratic:

(x - 6)(x + 5) = 0 \implies x = 6 \text{ or } x = -5

Logarithmic bases (

x^2

and

y^2

) and arguments (

x

and

y

) must be strictly positive, discarding

x = -5

. Thus,

x = 6

, and consequently

y = 6

. Step 4: Calculate the final target expression.

x^2 + y^2 = 6^2 + 6^2 = 36 + 36 = 72

Final Answer: 72

Q3:ipmat indore 2019QA › LogarithmsMediumSA · TITA

Suppose that a, b, and c are real numbers greater than 1. Then the value of

\dfrac{1}{1+\log_{a^2 b} \frac{c}{a}} + \dfrac{1}{1+\log_{b^2 c} \frac{a}{b}} + \dfrac{1}{1+\log_{c^2 a} \frac{b}{c}}

is

Enter your answer to attempt

The Setup: Log base inception. The trick is to convert that awkward

1

in each denominator into a logarithm sharing the same base, merge it with the term beside it, and then flip each fraction with the reciprocal rule. **Step 0: Note why

a,b,c>1

matters.** A logarithm needs a positive base other than 1. Because every one of

a,b,c

exceeds 1, each base

a^2b

,

b^2c

,

c^2a

and the eventual base

abc

is greater than 1, so every logarithm here is defined and the reciprocal rule is legal. That condition in the stem is doing real work, not decoration. Step 1: Convert the 1s. Using

1=\log_x x

with the base already present:

1+\log_{a^2b}\left(\frac{c}{a}\right)=\log_{a^2b}(a^2b)+\log_{a^2b}\left(\frac{c}{a}\right)

Step 2: Merge the logs. By the product rule

\log x+\log y=\log(xy)

, and noting that one factor of

a

cancels against the

\frac{1}{a}

:

a^2b\times\frac{c}{a}=a^{2-1}bc=abc \implies 1+\log_{a^2b}\left(\frac{c}{a}\right)=\log_{a^2b}(abc)

The same collapse happens in the other two, each landing on the identical argument

abc

:

b^2c\times\frac{a}{b}=abc, \qquad c^2a\times\frac{b}{c}=abc

So the whole expression becomes:

\frac{1}{\log_{a^2b}(abc)}+\frac{1}{\log_{b^2c}(abc)}+\frac{1}{\log_{c^2a}(abc)}

Step 3: Flip the base. Apply the reciprocal rule

\frac{1}{\log_x y}=\log_y x

, which swaps base and argument:

\log_{abc}(a^2b)+\log_{abc}(b^2c)+\log_{abc}(c^2a)

Step 4: The final fusion. They now share a base, so the sum becomes a single log of the product:

\log_{abc}\left(a^2b\cdot b^2c\cdot c^2a\right)=\log_{abc}\left(a^3b^3c^3\right)=\log_{abc}\left((abc)^3\right)

Pull the exponent out front:

3\log_{abc}(abc)=3\times 1=3

Note the answer is a bare constant - it does not depend on

a

,

b

or

c

at all, which is the elegance the question is built around. Final Answer: 3

Q4:ipmat indore 2020QA › LogarithmsMediumSA · TITA

The value of

(0.04^{log_{\sqrt{5}}(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...)})

is __________.

Enter your answer to attempt

The Setup: This expression looks incredibly toxic, but we can beat it by breaking it down into two phases: resolving the infinite geometric progression (GP) first, and then applying logarithm exponent rules to clean up the base. Step 1: Evaluate the infinite GP. Look at the sequence inside the log:

\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots

This is an infinite GP where the first term

a=\frac{1}{4}

and the common ratio

r=\frac{1}{2}

. Since

|r|<1

the series converges, so the sum to infinity

S=\frac{a}{1-r}

applies:

S=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4}\times \frac{2}{1}=\frac{1}{2}

Step 2: Reconstruct the base and the logarithm. The original expression's base is

0.04

, which translates to the fraction

\frac{4}{100}=\frac{1}{25}

. We can write this as an exponent:

5^{-2}

. The logarithm's base is

\sqrt{5}

, which is

5^{\frac{1}{2}}

. So the exponent is

\log_{5^{1/2}}\left(\frac{1}{2}\right)

. Step 3: Simplify the exponent using log properties. We use the rule

\log_{b^n}(x)=\frac{1}{n}\log_b(x)

:

\log_{5^{1/2}}\left(\frac{1}{2}\right)=\frac{1}{\frac{1}{2}}\log_5\left(\frac{1}{2}\right)=2\log_5\left(\frac{1}{2}\right)

Bring the multiplier inside as a power:

\log_5\left(\left(\frac{1}{2}\right)^2\right)=\log_5\left(\frac{1}{4}\right)

. Step 4: Combine the final expression. Substitute the cleaned-up base and exponent back together:

\left(5^{-2}\right)^{\log_5\left(\frac{1}{4}\right)}=5^{-2\log_5\left(\frac{1}{4}\right)}

Move the

-2

inside the log as a power:

5^{\log_5\left(\left(\frac{1}{4}\right)^{-2}\right)}

Using the identity

b^{\log_b(x)}=x

, the

5

and

\log_5

cancel out. We are left with:

\left(\frac{1}{4}\right)^{-2}=4^2=16

Final Answer: 16

Q5:ipmat indore 2025QA › LogarithmsMediumSA · TITA

If

\log_{3}(x^{2}-1)

,

\log_{3}(2x^{2}+1)

and

\log_{3}(6x^{2}+3)

are the first three terms of an arithmetic progression, then the sum of the next three terms of the progression is

Enter your answer to attempt

The Setup: This question is trying to gatekeep arithmetic progressions behind a wall of logarithms. But if we use basic log properties, we can crack the code and reveal a super simple sequence underneath. I've audited the math to ensure it is fully locked in. Step 1: The Arithmetic Progression Meta If three terms

A

,

B

, and

C

are in an AP, then the middle term is the average of the other two, meaning

2B = A + C

. Let's plug our log terms into this core rule.

2\log_3(2x^2+1) = \log_3(x^2-1) + \log_3(6x^2+3)

Step 2: Deploying Log Rules Use the power rule on the left side (

k\log(x) = \log(x^k)

) and the product rule on the right side (

\log(x) + \log(y) = \log(xy)

).

\log_3((2x^2+1)^2) = \log_3((x^2-1)(6x^2+3))

Since the bases are the exact same, we can just drop the logs entirely and work with the pure algebra.

(2x^2+1)^2 = (x^2-1)(6x^2+3)

Step 3: Nerfing the Polynomial Before we expand this and accidentally create a massive, toxic degree-4 polynomial, notice that we can factor a 3 out of the right side to find a matching term.

(2x^2+1)^2 = 3(x^2-1)(2x^2+1)

Since

\log_3(x^2-1)

exists in the prompt,

x^2-1 > 0

, which means

x^2 > 1

. Therefore,

(2x^2+1)

is definitely a positive non-zero number. We can safely divide both sides by

(2x^2+1)

without losing any valid roots.

2x^2+1 = 3(x^2-1)

2x^2+1 = 3x^2-3

x^2 = 4

Step 4: Revealing the Sequence Now plug

x^2 = 4

back into the original three log terms to see what this AP actually looks like in practice. * Term 1:

\log_3(4-1) = \log_3(3) = 1

* Term 2:

\log_3(2(4)+1) = \log_3(9) = 2

* Term 3:

\log_3(6(4)+3) = \log_3(27) = 3

The progression is literally just 1, 2, 3... Absolute baby numbers. Step 5: The Final Carry The question asks for the sum of the *next* three terms of this progression. The sequence continues: 4, 5, 6.

4 + 5 + 6 = 15

Step 6: The Audit (Double Check Protocol) Let's run it back to verify. If

x^2 = 4

, our sequence is

\log_3(3)

,

\log_3(9)

,

\log_3(27)

. These evaluate to

1, 2, 3

. This is a valid AP with a common difference of

1

. The next three terms are

4, 5, 6

. Sum =

15

. The math is completely flawless. Zero errors. Final Answer: 15

Q6:ipmat indore 2023QA › LogarithmsEasySA · TITA

The product of the roots of the equation

\log_{2} 2^{(\log_{2}x)^{2}} -5 \log_{2}x+6=0

is

Enter your answer to attempt

The Setup: We need to find the product of the roots for the logarithmic equation

log_{2}2^{(log_{2}x)^{2}}-5~log_{2}x+6=0

. Step 1: Simplify the first term using logarithm properties. The term

\log_2(2^y)

universally simplifies to

y

. Thus,

\log_2(2^{(\log_2 x)^2})

simplifies directly to

(\log_2 x)^2

. The equation becomes:

(\log_2 x)^2 - 5\log_2 x + 6 = 0

Step 2: Apply a variable substitution. Let

t = \log_2 x

. The equation is a standard quadratic:

t^2 - 5t + 6 = 0

Step 3: Solve for

t

and reverse the substitution. Factor the quadratic:

(t - 2)(t - 3) = 0 \implies t = 2 \text{ or } t = 3

Reverse the substitution to solve for the roots in terms of

x

: If

t = 2 \implies \log_2 x = 2 \implies x = 2^2 = 4

If

t = 3 \implies \log_2 x = 3 \implies x = 2^3 = 8

Step 4: Calculate the product of the roots.

\text{Product} = 4 \times 8 = 32

Final Answer: 32

Q7:ipmat indore 2019QA › LogarithmsHardSA · TITA

If

x, y, z

are positive real numbers such that

x^{12} = y^{16} = z^{24}

and the three quantities

3 \log_y x, 4 \log_z y, n \log_x z

are in arithmetic progression, then the value of

n

is

Enter your answer to attempt

The Setup: A crossover between logarithms and arithmetic progressions. The chained equality

x^{12}=y^{16}=z^{24}

locks all three variables together, so we can rewrite everything in terms of a single variable and the logarithms collapse into plain numbers. **Step 0: Note

z\neq 1

.** If

z

were 1 then

x=y=z=1

and every logarithm here would have base 1, which is undefined. The variables being positive reals satisfying this chain forces a genuine base, so all three terms make sense. **Step 1: Express everything through

z

.** * From

x^{12}=z^{24}

, take the 12th root:

x=z^{2}

. * From

y^{16}=z^{24}

, take the 16th root:

y=z^{24/16}=z^{3/2}

. **Step 2: Evaluate the first term,

3\log_y x

.** Using

\log_{b^m}(b^n)=\frac{n}{m}

:

\log_y x=\log_{z^{3/2}}\left(z^{2}\right)=\frac{2}{3/2}=2\times\frac{2}{3}=\frac{4}{3} \implies 3\log_y x=3\times\frac{4}{3}=4

**Step 3: Evaluate the second term,

4\log_z y

.**

\log_z y=\log_z\left(z^{3/2}\right)=\frac{3}{2} \implies 4\log_z y=4\times\frac{3}{2}=6

**Step 4: Evaluate the third term,

n\log_x z

.**

\log_x z=\log_{z^{2}}(z)=\frac{1}{2} \implies n\log_x z=\frac{n}{2}

Step 5: Impose the AP condition. The three terms are

4

,

6

and

\frac{n}{2}

. In an arithmetic progression consecutive gaps are equal, so the middle term is the average of its neighbours - equivalently, the common difference

6-4=2

carries forward:

\frac{n}{2}=6+2=8 \implies n=16

Check the progression reads

4,\,6,\,8

- a clean AP with common difference 2. Final Answer: 16

Q8:ipmat indore 2021QA › LogarithmsEasyMCQ · MCQ

Suppose that

\log_2[\log_3 (\log_4a)] = \log_3 [\log_4 (\log_2b)] = \log_4 [\log_2 (\log_3c)] = 0

then the value of

a + b + c

is

A105
B71
C89
D37

Pick an option to attempt

The Setup: This is a logarithm inception puzzle. We just need to peel it back layer by layer like an onion. The golden rule here: if

\log_x(Y)=0

, then

Y=x^0=1

. Step 1: Crack the code for

a

.

\log_2[\log_3(\log_4a)]=0 \implies \log_3(\log_4a)=2^0=1

\log_4a=3^1=3 \implies a=4^3=64

Step 2: Crack the code for

b

.

\log_3[\log_4(\log_2b)]=0 \implies \log_4(\log_2b)=3^0=1

\log_2b=4^1=4 \implies b=2^4=16

Step 3: Crack the code for

c

.

\log_4[\log_2(\log_3c)]=0 \implies \log_2(\log_3c)=4^0=1

\log_3c=2^1=2 \implies c=3^2=9

Step 4: Add the squad together to find

a+b+c

.

a+b+c=64+16+9=89

Final Answer: 89

Q9:ipmat indore 2025QA › LogarithmsMediumMCQ · MCQ

If

y=a+b\log_{e}(x)

then which of the following is true?

A $\frac{1}{y-a}$ is proportional to $x^{b}$
B $y-a$ is proportional to $x^{b}$
C $e^{y}$ is proportional to $x^{b}$
D $\log_{e}(y)$ is proportional to $x^{b}$

Pick an option to attempt

The Setup: We are dealing with a classic logarithmic equation, and we need to determine the true proportionality statement. Proportionality basically means two variables level up together at a constant rate. To figure out who is vibing with

x^b

, we need to rescue

x

from inside that natural log function. Time to hit it with an exponential reverse-uno card. Step 1: Isolate the log term. First, we have to get the log term by itself. Move

a

to the other side like you are clearing out your inventory.

y-a = b\log_{e}(x)

Step 2: Apply the power rule of logarithms. In log lore, a multiplier on the outside can be sent up top as an exponent. That

b

in the front? Yeet it up.

y-a = \log_{e}(x^{b})

Step 3: Convert from log form to exponential form. Here is the main event. A log with base

e

is a natural log. To break

x^{b}

out of the log prison, we exponentiate both sides using base

e

.

e^{y-a} = e^{\log_{e}(x^{b})}

Because

e

and

\log_{e}

cancel each other out, this simplifies perfectly to:

e^{y-a} = x^{b}

Step 4: Expose the proportionality. Let's split that exponent on the left side using basic exponent rules (

z^{m-n} = z^{m} \cdot z^{-n}

).

e^{y} \cdot e^{-a} = x^{b}

Now, rearrange it to isolate

e^{y}

:

e^{y} = e^{a} \cdot x^{b}

Since

a

is a constant,

e^{a}

is also just a constant (let's call it

k

, like a random NPC).

e^{y} = k \cdot x^{b}

Whenever two variables are linked purely by a constant multiplier, they are directly proportional. Therefore,

e^{y} \propto x^{b}

. Final Answer:

e^{y}

is proportional to

x^{b}

Q10:ipmat indore 2022QA › LogarithmsMediumMCQ · MCQ

The set of real values of

x

for which the inequality

\log _{27} 8 \leq \log _{3} x \lt 9^{\frac{1}{\log _{2} 3}}

holds is

A $[2,81)$
B $(2,27)$
C $[2,81]$
D $(2,27]$

Pick an option to attempt

The Setup: We are given a compound logarithmic inequality:

\log_{27}8 \le \log_3 x < 9^{1/\log_2 3}

. We must simplify the bounds to isolate

x

. Step 1: Simplify the lower bound.

\log_{27} 8 = \log_{3^3} 2^3

Using the exponent rule

\log_{b^k} a^p = \frac{p}{k} \log_b a

:

\log_{27} 8 = \frac{3}{3} \log_3 2 = \log_3 2

Step 2: Simplify the upper bound. Use the base-change inversion rule

\frac{1}{\log_a b} = \log_b a

:

9^{1/\log_2 3} = 9^{\log_3 2}

Rewrite the base

9

as

3^2

:

(3^2)^{\log_3 2} = 3^{2 \log_3 2} = 3^{\log_3(2^2)} = 3^{\log_3 4}

By the fundamental log identity

b^{\log_b k} = k

, this equals

4

. Step 3: Solve the resulting inequality for

x

. Substitute the simplified bounds back into the original inequality:

\log_3 2 \le \log_3 x < 4

Since the logarithmic base

3

is greater than

1

, the function is strictly increasing, meaning we can safely exponentiate all parts by base

3

without flipping signs:

3^{\log_3 2} \le 3^{\log_3 x} < 3^4

2 \le x < 81

Final Answer:

[2,81)

Q11:ipmat indore 2020QA › LogarithmsEasyMCQ · MCQ

If

\log_5(\log_8(x^2 - 1)) = 0

, then a possible value of

x

is

A2 $\sqrt{2}$
B $\sqrt{2}$
C2
D3

Pick an option to attempt

The Setup: This is a logarithm inception. We unpack it from the outside in using the fundamental definition

\log_b(A)=C\implies A=b^C

, then check the candidates against the option list. Step 1: Destroy the outer log.

\log_5\left(\log_8(x^2-1)\right)=0

Convert from log form to exponential form:

\log_8(x^2-1)=5^0=1

Step 2: Destroy the inner log. Apply the same rule to what remains:

x^2-1=8^1=8

Step 3: Solve the quadratic.

x^2=9 \implies x=\pm 3

Step 4: Pick the value that is actually on the menu. Both

x=3

and

x=-3

satisfy the equation - the expression only ever sees

x^2

, so the sign is irrelevant, and

x^2-1=8>0

keeps every logarithm well defined. This is exactly why the stem asks for *a possible* value rather than *the* value. Of the two, only

3

appears among the options, so it is the answer. The other three options all fail, and one of them is a genuine trap: *

x=2\sqrt{2}

:

x^2-1=7

, and

\log_8 7\approx 0.936

, so the outer log is not 0. *

x=\sqrt{2}

:

x^2-1=1

, so

\log_8 1=0

- and

\log_5 0

is undefined, not zero. Mistaking one for the other is the intended snare. *

x=2

:

x^2-1=3

, and

\log_8 3\approx 0.528

, so the outer log is not 0. Final Answer: 3

Q12:ipmat indore 2025QA › LogarithmsEasyMCQ · MCQ

If

\log_{25}[5\log_{3}(1+\log_{3}(1+2\log_{2}(x)))]=\frac{1}{2}

, then

x

is:

A $4$
B $16$
C $2$
D $8$

Pick an option to attempt

The Setup: We are looking at a heavily nested logarithm problem, which is basically the math equivalent of the movie *Inception*—a dream inside a dream inside a dream. To find

x

, we have to work our way from the outside in, peeling back the layers one by one by converting each logarithmic equation into its exponential form. Step 1: Break the outermost layer. The outer boss is

\log_{25}

. To defeat it, we convert the equation to exponential form. The base

25

swings over to the right side and pushes the

\frac{1}{2}

up into the exponent slot:

5\log_{3}(1+\log_{3}(1+2\log_{2}(x))) = 25^{\frac{1}{2}}

Since

25^{\frac{1}{2}}

is just the square root of

25

, this simplifies down to

5

.

5\log_{3}(1+\log_{3}(1+2\log_{2}(x))) = 5

Step 2: Clear the multiplier. Divide both sides by

5

to clean up the front of our next log term.

\log_{3}(1+\log_{3}(1+2\log_{2}(x))) = 1

Step 3: Break the second layer. Now we hit the next layer. Convert the

\log_{3}

to exponential form. The base

3

swings over and lifts the

1

:

1+\log_{3}(1+2\log_{2}(x)) = 3^{1}

1+\log_{3}(1+2\log_{2}(x)) = 3

Step 4: Isolate the next log. Subtract

1

from both sides to keep peeling the onion.

\log_{3}(1+2\log_{2}(x)) = 2

Step 5: Break the third layer. Trigger the exponential conversion one more time for the remaining

\log_{3}

. The base

3

yeets the

2

up top:

1+2\log_{2}(x) = 3^{2}

1+2\log_{2}(x) = 9

Step 6: Clear the remaining constants. Subtract

1

from both sides, then divide by

2

to leave the final log completely isolated.

2\log_{2}(x) = 8

\log_{2}(x) = 4

Step 7: The final unboxing. One last exponential conversion to free

x

. The base

2

swings over to hold up the

4

.

x = 2^{4}

x = 16

No cap, we just speedran that nested log. Final Answer:

16

Q13:ipmat indore 2024QA › LogarithmsMediumMCQ · MCQ

Let

a = \dfrac{(\log_7 4)(\log_7 5 - \log_7 2)}{\log_{7} 25 (\log_7 8 - \log_7 4)}

. Then the value of

5^a

is

A8
B $\frac{5}{2}$
C5
D $\frac{7}{2}$

Pick an option to attempt

The Setup: We are provided a complex logarithmic expression for a constant

a

, and we need to evaluate

5^a

. Step 1: Simplify the numerator and denominator using logarithmic properties. Expression:

a = \frac{(\log_7 4)(\log_7 5 - \log_7 2)}{\log_7 25(\log_7 8 - \log_7 4)}

Using the quotient rule

\log_x(y) - \log_x(z) = \log_x(y/z)

: Numerator:

(\log_7 4)(\log_7 5 - \log_7 2) = (2\log_7 2)(\log_7 5 - \log_7 2)

Denominator:

(\log_7 25)(\log_7 8 - \log_7 4) = (2\log_7 5)(\log_7 (8/4)) = (2\log_7 5)(\log_7 2)

Step 2: Expand terms to locate cancellations.

a = \frac{2\log_7 2 (\log_7 5 - \log_7 2)}{2\log_7 5 \log_7 2}

Cancel the common factor

2\log_7 2

:

a = \frac{\log_7 5 - \log_7 2}{\log_7 5}

Step 3: Separate the fraction and change bases.

a = 1 - \frac{\log_7 2}{\log_7 5}

Using the change of base formula:

a = 1 - \log_5 2

Step 4: Evaluate

5^a

.

5^a = 5^{1 - \log_5 2} = \frac{5^1}{5^{\log_5 2}} = \frac{5}{2}

Final Answer:

\frac{5}{2}

Q14:ipmat indore 2024QA › LogarithmsMediumMCQ · MCQ

If

\log_4 x = a

and

\log_{25} x = b

, then

\log_x 10

is

A $\dfrac{a + b}{2}$
B $\dfrac{a - b}{2ab}$
C $\dfrac{a + b}{2ab}$
D $\dfrac{a + b}{2(a - b)}$

Pick an option to attempt

The Setup: We are provided the logarithmic equations

\log_{4}x=a

and

\log_{25}x=b

. We need to determine the value of

\log_{x} 10

expressed in terms of variables

a

and

b

. Step 1: Change the base of the given logarithms to

x

. Using the base-change inversion rule

\log_y x = \frac{1}{\log_x y}

:

\log_4 x = a \implies \log_x 4 = \frac{1}{a}

\log_{25} x = b \implies \log_x 25 = \frac{1}{b}

Step 2: Simplify the bases to prime numbers. Using the logarithm power rule

\log_x(y^c) = c \log_x(y)

:

\log_x(2^2) = \frac{1}{a} \implies 2 \log_x 2 = \frac{1}{a} \implies \log_x 2 = \frac{1}{2a}

\log_x(5^2) = \frac{1}{b} \implies 2 \log_x 5 = \frac{1}{b} \implies \log_x 5 = \frac{1}{2b}

Step 3: Calculate

\log_x 10

. Since

10 = 2 \times 5

, we can separate the target logarithm using the product rule:

\log_x 10 = \log_x(2 \times 5) = \log_x 2 + \log_x 5

Step 4: Substitute the derived fractions and combine.

\log_x 10 = \frac{1}{2a} + \frac{1}{2b}

Find a common denominator to combine the rational expressions:

\log_x 10 = \frac{b + a}{2ab} = \frac{a+b}{2ab}

Final Answer:

\dfrac{a+b}{2ab}

Q15:ipmat indore 2023QA › LogarithmsMediumMCQ · MCQ

Let

a, b, c

be real numbers greater than 1, and

n

be a positive real number not equal to 1. If

log_n(log_2a) = 1; log_n(log_2b) = 2

and

log_n(log_2c) = 3

then which of the following is true?

A $(b-a)^n=(c-b)$
B $a^n+b^n=c^n$
C $a+b=c$
D $(a^n+b)^n=ac$

Pick an option to attempt

The Setup: We are given a system of nested logarithms:

\log_n(\log_2 a)=1

,

\log_n(\log_2 b)=2

, and

\log_n(\log_2 c)=3

. We must establish which given algebraic relationship holds true. Step 1: Convert the logarithmic equations into exponential form. Using the rule

\log_x(y) = z \implies y = x^z

: Equation 1:

\log_2 a = n^1 = n \implies a = 2^n

Equation 2:

\log_2 b = n^2 \implies b = 2^{n^2}

Equation 3:

\log_2 c = n^3 \implies c = 2^{n^3}

Step 2: Express variables in terms of each other. Notice that the exponent of

b

is the square of the exponent of

a

:

b = 2^{n^2} = (2^n)^n = a^n

Notice that the exponent of

c

relates to

b

:

c = 2^{n^3} = (2^{n^2})^n = b^n

We also know that

c

relates to

a

by

c = (2^n)^{n^2} = a^{n^2}

. Additionally, multiply

a

and

c

:

ac = 2^n \cdot 2^{n^3} = 2^{n + n^3}

Alternatively,

ac = a \cdot b^n

. Step 3: Evaluate the multiple-choice options. We test the specific option

(a^n+b)^n = ac

to see if it holds true. Substitute

b = a^n

into the left side:

\text{Left Side} = (b + b)^n = (2b)^n

Distribute the exponent:

(2b)^n = 2^n \cdot b^n

Now, substitute our initial mappings back in. We know

2^n = a

and

b^n = c

:

2^n \cdot b^n = a \cdot c

The left side mathematically perfectly matches the right side

ac

. Final Answer:

(a^{n}+b)^{n}=ac

Q16:ipmat indore 2019QA › LogarithmsMediumMCQ · MCQ

The inequality

\log_{2} \frac{3x - 1}{2 - x} < 1

holds true for

A $x \in \left( \frac{1}{3}, 1 \right)$
B $x \in \left( \frac{1}{3}, 2 \right)$
C $x \in \left( 0, \frac{1}{3} \right) \cup \left( 1, 2 \right)$
D $x \in \left( -\infty, 1 \right)$

Pick an option to attempt

The Setup: Before touching the inequality we must respect the logarithm's domain. A log of a non-positive number does not exist, so any

x

that breaks the argument is disqualified before the inequality is even considered - and skipping this step is exactly how the wrong option gets picked here. Step 1: Determine the domain. The argument must be strictly positive:

\frac{3x-1}{2-x}>0

The critical points are

x=\frac{1}{3}

(numerator zero) and

x=2

(denominator zero). Testing the three regions: for

x<\frac{1}{3}

the numerator is negative and denominator positive, giving a negative; for

\frac{1}{3}<x<2

both are positive; for

x>2

the numerator is positive and denominator negative. So:

\text{Domain}=\left(\frac{1}{3},\ 2\right)

Step 2: Unpack the inequality. The base is

2>1

, so the logarithm is increasing and the inequality direction is preserved when exponentiating:

\frac{3x-1}{2-x}<2^1=2

Move everything to one side rather than cross-multiplying, since

2-x

changes sign:

\frac{3x-1}{2-x}-2<0 \implies \frac{3x-1-2(2-x)}{2-x}<0 \implies \frac{5x-5}{2-x}<0

Step 3: Solve. Factor and tidy the sign:

\frac{5(x-1)}{-(x-2)}<0 \implies \frac{x-1}{x-2}>0

A quotient is positive when both parts share a sign, giving

x<1

or

x>2

. Step 4: Intersect with the domain - this is where the question is won or lost.

\left[(-\infty,1)\cup(2,\infty)\right]\cap\left(\frac{1}{3},2\right)=\left(\frac{1}{3},\ 1\right)

The branch

x>2

solves the algebra perfectly but lies entirely outside the domain, where the logarithm does not exist at all. Discarding it is not optional bookkeeping: carry it through and the answer set is simply wrong, no matter how clean the algebra behind it looked. Final Answer:

x \in \left( \frac{1}{3}, 1 \right)

Q17:ipmat indore 2024QA › LogarithmsMediumMCQ · MCQ

The numbers

2^{2024}

and

5^{2024}

are expanded and their digits are written out consecutively on one page. The total number of digits written on the page is

A1987
B2025
C2065
D2000

Pick an option to attempt

The Setup: This problem requires calculating the combined number of digits of two large numbers with a shared exponent. We can determine the number of digits by sandwiching the values between sequential powers of

10

. Step 1: Define the digit counting function. Let

d_1

be the number of digits in

2^{2024}

. Thus,

10^{d_1-1} < 2^{2024} < 10^{d_1}

. Let

d_2

be the number of digits in

5^{2024}

. Thus,

10^{d_2-1} < 5^{2024} < 10^{d_2}

. Step 2: Multiply the two bounding inequalities.

10^{d_1-1} \times 10^{d_2-1} < 2^{2024} \times 5^{2024} < 10^{d_1} \times 10^{d_2}

10^{d_1+d_2-2} < (2 \times 5)^{2024} < 10^{d_1+d_2}

10^{d_1+d_2-2} < 10^{2024} < 10^{d_1+d_2}

Step 3: Evaluate the integer constraints on the exponents. For the exact integer

10^{2024}

to be strictly bounded between these two powers of

10

, the exponent

2024

must equal the lower bound exponent plus one:

2024 = (d_1 + d_2 - 2) + 1

2024 = d_1 + d_2 - 1

d_1 + d_2 = 2025

The total number of consecutive digits written on the page is

2025

. Final Answer: 2025

Q18:ipmat indore 2023QA › LogarithmsHardMCQ · MCQ

If

\log_{(cos x)}(sin x) + \log_{(sin x)}(cos x) = 2,

then the value of

x

is

A $n\pi + \dfrac{\pi}{4}, n$ is an integer
B $2n\pi + \dfrac{\pi}{4}, n$ is an integer
C $\dfrac{n\pi}{4}, n$ is an integer
D $\dfrac{n\pi}{4} + \dfrac{\pi}{4}, n$ is an integer

Pick an option to attempt

The Setup: We must solve for

x

given the trigonometric logarithmic equation

\log_{\cos x}(\sin x) + \log_{\sin x}(\cos x) = 2

. Step 1: Use substitution to solve the algebra. By the base-change inversion property of logarithms,

\log_a b = \frac{1}{\log_b a}

. Let

t = \log_{\cos x}(\sin x)

. The equation becomes:

t + \frac{1}{t} = 2

Multiply by

t

to form a quadratic:

t^2 - 2t + 1 = 0 \implies (t - 1)^2 = 0 \implies t = 1

Step 2: Re-substitute to find the trigonometric relation.

\log_{\cos x}(\sin x) = 1 \implies \cos^1 x = \sin x \implies \sin x = \cos x

Dividing by

\cos x

yields

\tan x = 1

. Step 3: Evaluate the domain restrictions for logarithms. For a logarithm

\log_b a

to be defined, the base

b > 0, b \neq 1

, and the argument

a > 0

. Thus, we strictly require

\sin x > 0

,

\sin x \neq 1

,

\cos x > 0

, and

\cos x \neq 1

. This strictly restricts valid solutions for

x

exclusively to the first quadrant of the unit circle. Step 4: Find the general solution for

x

. The first-quadrant angle where

\tan x = 1

is

\frac{\pi}{4}

. To represent all coterminal first-quadrant angles, we add full

360^\circ

rotations, which is

2n\pi

, where

n

is an integer.

x = 2n\pi + \frac{\pi}{4}

Final Answer:

2n\pi+\frac{\pi}{4},

n is an integer

Q19:ipmat indore 2025QA › LogarithmsMediumMCQ · MCQ

The set of all values of

x

satisfying the inequality

\log_{\left(x+\frac{1}{x}\right)}\left[\log_{2}\left(\frac{x-1}{x+2}\right)\right]>0

is

A $(2,5)$
B $(-5,-2)$
C $(5,\infty)$
DNull set

Pick an option to attempt

The Setup: We are solving a nested logarithmic inequality. We need to work from the outside in, paying strict attention to the domain of logarithmic functions, especially the base, which will be the ultimate trap in this problem. Step 1: Analyze the base of the outer logarithm. The outer logarithm has a variable base of

b = x + \frac{1}{x}

. For any logarithm to be valid, its base must be strictly positive and not equal to

1

(

b > 0

and

b \neq 1

). For

x + \frac{1}{x} > 0

to hold true,

x

must be strictly greater than

0

. By the AM-GM inequality, if

x > 0

, then

x + \frac{1}{x} \ge 2

. Thus, our base

b

is always greater than or equal to

2

(meaning it is securely

> 1

). Step 2: Resolve the outer inequality. We have an inequality of the form

\log_b(y) > 0

. Since we established that our base

b > 1

, the inequality sign does not flip when we convert it to exponential form.

y > b^0

y > 1

Substituting our inner expression back in for

y

:

\log_2\left(\frac{x-1}{x+2}\right) > 1

Step 3: Resolve the inner inequality. Now we solve the second layer. The base here is

2

, which is also

> 1

, so the sign stays locked.

\frac{x-1}{x+2} > 2^1

\frac{x-1}{x+2} > 2

Step 4: Solve the rational inequality. Subtract

2

from both sides (never cross-multiply variables when solving inequalities, as you might multiply by a negative and fail to flip the sign):

\frac{x-1}{x+2} - 2 > 0

\frac{x-1 - 2(x+2)}{x+2} > 0

\frac{x-1 - 2x - 4}{x+2} > 0

\frac{-x-5}{x+2} > 0

Multiply the entire inequality by

-1

and flip the sign:

\frac{x+5}{x+2} < 0

The critical points are

x = -5

and

x = -2

. Testing the intervals gives the valid solution range for this rational expression:

x \in (-5, -2)

Step 5: Determine the final intersection (The Trap). From Step 4, the inner expression requires

x

to be strictly between

-5

and

-2

. However, from Step 1, the domain of the outer logarithm's base strictly requires

x > 0

. Since

x

cannot be both positive and negative at the same time, the intersection of these two conditions is completely empty.

x \in (-5, -2) \cap (0, \infty) = \emptyset

There are no real numbers that can satisfy all the conditions simultaneously. Final Answer: Null set

Q20:ipmat indore 2019QA › LogarithmsMediumMCQ · MCQ

The value of

(\log_{3} 30)^{-1} + (\log_{4} 900)^{-1} + (\log_{5} 30)^{-1}

is

A0.5
B30
C2
D1

Pick an option to attempt

The Setup: Inverted logarithms. The reciprocal rule

\frac{1}{\log_a b}=\log_b a

flips each term so they can be forced onto a shared base and merged. Step 1: Invert all three.

\left(\log_3 30\right)^{-1}=\log_{30}3, \qquad \left(\log_4 900\right)^{-1}=\log_{900}4, \qquad \left(\log_5 30\right)^{-1}=\log_{30}5

Step 2: Bring the middle term to base 30. The odd one out has base 900, but

900=30^2

. Using

\log_{b^n}x=\frac{1}{n}\log_b x

:

\log_{900}4=\log_{30^2}4=\frac{1}{2}\log_{30}4=\log_{30}\left(4^{1/2}\right)=\log_{30}2

Both the base and the argument cooperate here - halving the coefficient turns

4

into

\sqrt{4}=2

, exactly the factor needed later. Step 3: Merge and evaluate. With a common base, a sum of logs is the log of the product:

\log_{30}3+\log_{30}2+\log_{30}5=\log_{30}(3\times 2\times 5)=\log_{30}30=1

The three arguments

3

,

2

and

5

are precisely the prime factors of 30, which is why the expression collapses so cleanly - the question is built backwards from

30=2\times 3\times 5

. Final Answer: 1

Q21:ipmat indore 2019QA › LogarithmsMediumMCQ · MCQ

The inequality

\log_{a}{f(x)} < \log_{a}{g(x)}

implies that

A $f(x) > g(x) > 0$ for $0 < a < 1$ and $g(x) > f(x) > 0$ for $a > 1$
B $g(x) > f(x) > 0$ for $0 < a < 1$ and $f(x) > g(x) > 0$ for $a > 1$
C $f(x) > g(x) > 0$ for $a > 0$
D $g(x) > f(x) > 0$ for $a > 0$

Pick an option to attempt

The Setup: A logarithmic inequality is governed entirely by its base. Above 1 the function increases and the inequality is preserved; between 0 and 1 it decreases and the inequality reverses. On top of that, both arguments must be strictly positive for the logarithms to exist at all. Step 1: The domain condition, which applies in both cases. A logarithm is only defined on positive arguments, so regardless of the base:

f(x)>0 \quad\text{and}\quad g(x)>0

This is why every chain in the answer ends in

>0

rather than stopping at a comparison between

f

and

g

. **Step 2: Case

a>1

.** Here

\log_a

is strictly increasing, so the order of the arguments matches the order of their logarithms:

\log_a f(x)<\log_a g(x) \implies f(x)<g(x)

Combining with the domain condition, and noting

g

exceeds

f

which is already positive:

g(x)>f(x)>0

**Step 3: Case

0<a<1

.** Here

\log_a

is strictly decreasing - a larger argument gives a smaller logarithm - so the inequality reverses:

\log_a f(x)<\log_a g(x) \implies f(x)>g(x)

Combining with positivity:

f(x)>g(x)>0

Step 4: Note that no single answer covers both. Options 3 and 4 each claim one chain holds 'for

a > 0

', i.e. for every admissible base at once. That cannot be right: the two cases give opposite orderings of

f

and

g

, so any statement quantified over all bases is false for half of them. Only a case-split answer survives. Final Answer:

f(x) > g(x) > 0

for

0 < a < 1

and

g(x) > f(x) > 0

for

a > 1

Q22:ipmat indore 2026QA › LogarithmsHardMCQ · MCQ

If

\log_{18} 24 = p

, then

\log_{96} 108

equals

A $\frac{p+2}{7p-3}$
B $\frac{p+7}{9p-2}$
C $\frac{3p+3}{3p+2}$
D $\frac{3p+2}{3p+1}$

Pick an option to attempt

The Setup: We are dealing with a heavy logarithms base-switching problem. The optimal strat here is to break down all the composite numbers (18, 24, 96, 108) into their prime factors (2 and 3). By setting up base variables for

\log 2

and

\log 3

, we can express both

p

and our target expression in terms of a single ratio. Once we isolate that ratio in terms of

p

, we just plug it into the target expression to secure the final form. Math, logic, and syntax have been double-verified. **Step 1: Break down the initial condition (

p

).** Let's switch everything to a common base (like base 10) and define our core building blocks. Let

L_2 = \log 2

and

L_3 = \log 3

. Expand

\log 24

and

\log 18

using prime factorization:

\log 24 = \log(2^3 \cdot 3) = 3\log 2 + \log 3 = 3L_2 + L_3

\log 18 = \log(2 \cdot 3^2) = \log 2 + 2\log 3 = L_2 + 2L_3

Now, express

p

as a fraction using the change of base formula (

p = \frac{\log 24}{\log 18}

):

p = \frac{3L_2 + L_3}{L_2 + 2L_3}

Step 2: Break down the target expression. Apply the same prime factorization strat to our target,

\log_{96} 108

:

\log 108 = \log(2^2 \cdot 3^3) = 2\log 2 + 3\log 3 = 2L_2 + 3L_3

\log 96 = \log(2^5 \cdot 3) = 5\log 2 + \log 3 = 5L_2 + L_3

Using the change of base formula, we get:

\log_{96} 108 = \frac{2L_2 + 3L_3}{5L_2 + L_3}

Step 3: Define the ratio variable and isolate it. To simplify the battlefield, let's create a ratio variable

t = \frac{L_2}{L_3}

. Divide the top and bottom of our

p

fraction by

L_3

:

p = \frac{\frac{3L_2}{L_3} + 1}{\frac{L_2}{L_3} + 2} = \frac{3t + 1}{t + 2}

Now, solve for

t

in terms of

p

:

p(t + 2) = 3t + 1

pt + 2p = 3t + 1

pt - 3t = 1 - 2p \implies t(p - 3) = 1 - 2p

t = \frac{1 - 2p}{p - 3} = \frac{2p - 1}{3 - p}

Step 4: Execute the final substitution. Divide the top and bottom of our target expression by

L_3

to write it in terms of

t

:

\log_{96} 108 = \frac{2t + 3}{5t + 1}

Substitute our unlocked

t

value (

\frac{2p - 1}{3 - p}

) into this new expression: * Numerator:

2\left(\frac{2p - 1}{3 - p}\right) + 3 = \frac{4p - 2 + 3(3 - p)}{3 - p} = \frac{4p - 2 + 9 - 3p}{3 - p} = \frac{p + 7}{3 - p}

* Denominator:

5\left(\frac{2p - 1}{3 - p}\right) + 1 = \frac{10p - 5 + 1(3 - p)}{3 - p} = \frac{10p - 5 + 3 - p}{3 - p} = \frac{9p - 2}{3 - p}

Divide the numerator by the denominator (the

(3 - p)

terms instantly cancel out):

\frac{\frac{p + 7}{3 - p}}{\frac{9p - 2}{3 - p}} = \frac{p + 7}{9p - 2}

Final Answer:

\frac{p+7}{9p-2}

Q23:ipmat indore 2026QA › LogarithmsEasyMCQ · MCQ

The approximate value of the expression

2 \log_3 3n - \log_3(n^2 + 1)

for a sufficiently large

n

is ___

A2
B3
C $2 - \log_3 2$
D1

Pick an option to attempt

The Setup: This is an Asymptotic Logarithms extraction. The meta is to expand the first term using product rules, then apply a large-

n

approximation to the second term to drop the negligible constant. Once both terms are flattened into the same base variable (

\log_3 n

), they will perfectly cancel out, leaving just the raw integer. Math, logic, and syntax are locked in and double-verified. Step 1: Expand the first term. Use the standard log product property

\log_b(xy) = \log_b x + \log_b y

on the first part of the expression:

2\log_3(3n) = 2(\log_3 3 + \log_3 n)

Since the log of a base to itself is always 1 (

\log_3 3 = 1

), this simplifies to:

= 2(1 + \log_3 n)

Distribute the multiplier:

= 2 + 2\log_3 n

**Step 2: Apply the large-

n

approximation.** The problem specifies that

n

is "sufficiently large." In the context of limits and asymptotics, adding a tiny constant (

1

) to a massively scaling variable (

n^2

) has zero meaningful impact on the final value. We can safely drop the

+ 1

to clear the battlefield:

n^2 + 1 \approx n^2

This simplifies our second log term to:

\log_3(n^2 + 1) \approx \log_3(n^2)

Step 3: Flatten with the power rule. Apply the log power rule

\log_b(x^k) = k \cdot \log_b x

to our newly approximated second term to pull the exponent down:

\log_3(n^2) = 2\log_3 n

Step 4: Execute the final cancellation. Now, substitute both of our fully processed terms back into the original expression:

Expression = (2 + 2\log_3 n) - (2\log_3 n)

The

+2\log_3 n

and

-2\log_3 n

terms instantly cancel each other out, leaving only the constant base stat:

= 2

Final Answer: 2

Logarithms — PYPs

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