Logarithms — PYPs

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

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

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

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

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

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

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

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

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 $x^{b}$

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: $[2,81)$

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

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: $16$

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