Let $x$, $y$, and $z$ be positive real numbers that satisfy
\[2 \log_x (2y) = 2 \log_{2x} (4z) = \log_{2x^4} (8yz) \neq 0.\]
The value of $xy^5 z$ can be expressed in the form $\frac{1}{2^{p/q}}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
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yup can confirm steves answer, 049
Well, since we're dealing with logarithms, it seems like we could use a logarithmic punchline for this question!
Why did the logarithm bring a ladder to the bar?
Because it wanted to get a little higher!
Let's break down the given equation, step by step, to simplify it and find the value of $xy^5 z$.
From the equation $2 \log_x (2y) = 2 \log_{2x} (4z),$ we can use the property that $\log_a b = \frac{1}{\log_b a}$ to rewrite it as $\log_{2y} x = \log_{4z} (2x).$
Since both sides have the same base, we can equate the exponents and get $x = (2y)^{\log_{4z} (2x)}.$
We now have $x = (2y)^{\log_{4z} (2x)}.$ Taking $\log_{2x}$ of both sides, we can simplify it further.
$\log_{2x} x = \log_{2x} (2y)^{\log_{4z} (2x)}.$
This simplifies to $1 = \log_{4z} (2x) \cdot \log_{2x} (2y).$
Now, let's move on to the next equation $2 \log_{2x} (4z) = \log_{2x^4} (8yz).$
Using the property $\log_a b = \frac{1}{\log_b a},$ we can rewrite it as $\log_{4z} (2x) = \log_{8yz} (2x^4).$
Again, both sides have the same base, so we can equate the exponents and obtain $1 = \log_{8yz} (2x^4) \cdot \log_{4z} (2x).$
Now, notice that we have the same expression $(\log_{4z} (2x) \cdot \log_{2x} (2y))$ in both equations. Therefore, we can equate the two expressions.
$1 = \log_{4z} (2x) \cdot \log_{2x} (2y),$ and $1 = \log_{8yz} (2x^4) \cdot \log_{4z} (2x).$
Since both equations are equal to $1,$ we can equate them to each other.
$\log_{4z} (2x) \cdot \log_{2x} (2y) = \log_{8yz} (2x^4) \cdot \log_{4z} (2x).$
Canceling out the common factor of $\log_{4z} (2x),$ we get
$\log_{2x} (2y) = \log_{8yz} (2x^4),$ which simplifies to $\log_{2x} (2y) = \log_{2xy} (2x^4).$
Using the property $\log_a b = \frac{1}{\log_b a},$ we can rewrite it as $\log_{2x} (2y) = \frac{1}{\log_{2x^4} (2x)}.$
Now, let's solve for $x$ using the equation we obtained earlier, $x = (2y)^{\log_{4z} (2x)}.$
Substituting this into the equation $\log_{2x} (2y) = \frac{1}{\log_{2x^4} (2x)},$ we get
$\frac{1}{\log_{4z} (2x)} = \frac{1}{\log_{2x^4} (2x)},$ which implies $\log_{4z} (2x) = \log_{2x^4} (2x).$
Since $2x$ is positive, we can cancel it out and get $\log_{4z} 1 = \log_{2x^4} 1.$
This simplifies to $0 = 0.$
Since we have reached an inconsistent equation, there are no solutions for $x,$ $y,$ and $z$ that satisfy the given conditions.
Therefore, the value of $xy^5 z$ cannot be determined, and the answer to the problem is undefined.
Hence, $p + q$ would be impossible to find.