To evaluate $f(x) = \log_2(x)$ at $x = 128$, we need to express 128 as a power of 2.

We can start by rewriting 128: $128 = 2^7$

Thus, we can substitute this into the logarithm: $f(128) = \log_2(128) = \log_2(2^7)$

Using the logarithmic property $\log_b(b^a) = a$, we have: $\log_2(2^7) = 7$

Therefore, $f(128) = 7.$

Summary:

$\log_2(128) = 7$

Would you like more details or have any questions? Here are some related questions you might consider:

1. How can you use logarithms to solve exponential equations?
2. What are some properties of logarithms that are useful in simplification?
3. How would you evaluate $\log_2(64)$?
4. Can you explain the change of base formula for logarithms?
5. What is the relationship between logarithms and exponential functions?

Tip: Remember that knowing the powers of 2 can help you quickly evaluate logarithms in base 2 without a calculator.