To solve $\log_2 81 = x$, we need to express this logarithmic equation in a form that we can solve for $x$.

We know that logarithms represent exponents. Specifically, the equation $\log_b a = x$ means $b^x = a$.

So, for $\log_2 81 = x$, this means:

$2^x = 81$

Next, we approximate $x$ by taking the logarithm of both sides with a more convenient base, such as base 10:

$\log_{10} (2^x) = \log_{10} 81$

Using the logarithmic identity $\log_b (a^n) = n \log_b a$, we get:

$x \log_{10} 2 = \log_{10} 81$

Solving for $x$:

$x = \frac{\log_{10} 81}{\log_{10} 2}$

Now, let's compute the values:

$x = \frac{\log_{10} 81}{\log_{10} 2} \approx \frac{1.9085}{0.3010} \approx 6.34$

Therefore, $x \approx 6.34$.

Would you like more details, or do you have any questions?

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Here are five related questions to deepen your understanding:

1. How do you solve logarithmic equations with different bases?
2. What are the logarithmic identities that simplify calculations?
3. How do you convert between logarithms of different bases?
4. How does the change-of-base formula work in logarithms?
5. What are some real-world applications of logarithms?

Tip: When using logarithms with different bases, the change-of-base formula $\log_b a = \frac{\log_c a}{\log_c b}$ is very helpful!