To solve $\log_{16} 64$, we need to express the logarithm in a solvable form.

We want to find $x$ such that:

$16^x = 64$

First, express both 16 and 64 as powers of 2:

• $16 = 2^4$
• $64 = 2^6$

Now substitute these values into the equation:

$(2^4)^x = 2^6$

Simplify:

$2^{4x} = 2^6$

Since the bases are the same, equate the exponents:

$4x = 6$

Solve for $x$:

$x = \frac{6}{4} = \frac{3}{2}$

Thus, $\log_{16} 64 = \frac{3}{2}$.

Would you like further details or have any questions?

Here are five related questions to expand on this concept:

1. What is $\log_{4} 64$?
2. Can you solve $\log_{64} 16$?
3. How would you calculate $\log_{32} 16$?
4. What is the value of $\log_{81} 9$?
5. How would you solve $\log_{16} 256$?

Tip: When working with logarithms, converting numbers to powers of a common base (e.g., powers of 2) often simplifies the problem.