Let's solve both parts (a) and (b) from the image using logarithmic properties and then verify the results using the change-of-base formula.

Part (a):

$\log_2(8)$

We know that $8 = 2^3$, so: $\log_2(8) = \log_2(2^3) = 3$ This is because $\log_b(b^x) = x$, where $b$ is the base and $x$ is the exponent.

Verification (Change-of-base formula):

The change-of-base formula is: $\log_b(a) = \frac{\log(a)}{\log(b)}$ For $\log_2(8)$: $\log_2(8) = \frac{\log(8)}{\log(2)}$ Since $\log(8) \approx 0.9031$ and $\log(2) \approx 0.3010$: $\frac{0.9031}{0.3010} \approx 3$ Thus, $\log_2(8) = 3$ is correct.

Part (b):

$\log_5 \left( \frac{1}{5} \right)$

We can express $\frac{1}{5}$ as $5^{-1}$, so: $\log_5 \left( \frac{1}{5} \right) = \log_5 \left( 5^{-1} \right) = -1$ This follows from the same property $\log_b(b^x) = x$.

Verification (Change-of-base formula):

Using the change-of-base formula again: $\log_5 \left( \frac{1}{5} \right) = \frac{\log \left( \frac{1}{5} \right)}{\log(5)}$ We know that $\log \left( \frac{1}{5} \right) = -\log(5)$, so: $\frac{-\log(5)}{\log(5)} = -1$ Thus, $\log_5 \left( \frac{1}{5} \right) = -1$ is correct.

Final Answers:

• (a) $\log_2(8) = 3$
• (b) $\log_5 \left( \frac{1}{5} \right) = -1$

Would you like more details or have any questions on these solutions?

Here are some related questions for practice:

1. Simplify $\log_2(16)$.
2. Solve $\log_3(27)$.
3. Simplify $\log_7 \left( \frac{1}{49} \right)$.
4. Use the change-of-base formula to compute $\log_4(64)$.
5. Solve $\log_9 \left( \frac{1}{81} \right)$.

Tip: Remember, if the base and the argument of the logarithm are the same (like $\log_b(b^x)$), the answer is always the exponent $x$.