We are given the equation:

$5 \log_2 x = -\log_2 32$

Step 1: Apply the properties of logarithms

We know that $\log_2 32$ is the logarithm of 32 with base 2. First, let's simplify $\log_2 32$.

Since $32 = 2^5$, we can write:

$\log_2 32 = \log_2 (2^5) = 5$

Thus, the equation becomes:

$5 \log_2 x = -5$

Step 2: Isolate $\log_2 x$

Divide both sides of the equation by 5 to isolate $\log_2 x$:

$\log_2 x = -1$

Step 3: Solve for $x$

Now, we rewrite the logarithmic equation in its exponential form. Recall that $\log_2 x = -1$ means:

$x = 2^{-1}$

Thus:

$x = \frac{1}{2}$

Final Answer:

$x = \frac{1}{2}$

Would you like a detailed explanation of any step, or do you have any questions?

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Here are 5 related questions you might want to explore:

1. What is the logarithmic form of the equation $\log_b x = y$?
2. How do you convert between logarithmic and exponential forms?
3. What are the properties of logarithms, such as product, quotient, and power rules?
4. How do you solve logarithmic equations with different bases?
5. What happens when you have logarithms on both sides of the equation?

Tip: When dealing with logarithms, always check for base transformations or simplifications like $\log_b b^n = n$ to make equations easier.