Let's solve each problem step-by-step.

3.

(a) $\log_5 125$

$125 = 5^3$ $\log_5 125 = \log_5 (5^3) = 3$

(b) $\log_5 \left(\frac{1}{125}\right)$

$\frac{1}{125} = 5^{-3}$ $\log_5 \left(\frac{1}{125}\right) = \log_5 (5^{-3}) = -3$

4.

(a) $\ln\left(\frac{1}{e}\right)$

$\frac{1}{e} = e^{-1}$ $\ln\left(\frac{1}{e}\right) = \ln(e^{-1}) = -1$

(b) $\log_{10} \sqrt{10}$

$\sqrt{10} = 10^{1/2}$ $\log_{10} \sqrt{10} = \log_{10} (10^{1/2}) = \frac{1}{2}$

5.

(a) $e^{4.5}$

$e^{4.5} \text{ is already in its simplest form.}$

(b) $\log_{10} 0.0001$

$0.0001 = 10^{-4}$ $\log_{10} 0.0001 = \log_{10} (10^{-4}) = -4$

6.

(a) $\log_5 2.25$

$2.25 = \left(\frac{3}{2}\right)^2$ $\log_5 2.25 = \log_5 \left(\left(\frac{3}{2}\right)^2\right) = 2 \log_5 \left(\frac{3}{2}\right)$

(b) $\log_4 500$

$\log_4 500 = \frac{\log_{10} 500}{\log_{10} 4}$

7.

(a) $\log_6 6 - \log_6 15 + \log_6 20$

$\log_6 6 = 1$ $\log_6 15 = \log_6 (3 \times 5) = \log_6 3 + \log_6 5$ $\log_6 20 = \log_6 (2^2 \times 5) = 2 \log_6 2 + \log_6 5$

Combining them: $1 - (\log_6 3 + \log_6 5) + (2 \log_6 2 + \log_6 5) = 1 - \log_6 3 + 2 \log_6 2$

(b) $\log_{100} 100 - \log_{100} 18 - \log_{100} 50$

$\log_{100} 100 = 1$ $\log_{100} 18 + \log_{100} 50 = \log_{100} (18 \times 50)$ $\log_{100} 900 = \log_{100} (100 \times 9) = \log_{100} 100 + \log_{100} 9 = 1 + \log_{100} 9$

Combining them: $1 - (1 + \log_{100} 9) = - \log_{100} 9$

8.

(a) $e^{2 + \ln 5}$

$e^{2 + \ln 5} = e^2 \times e^{\ln 5} = e^2 \times 5$

(b) $\ln \left(e^{e^4}\right)$

$\ln \left(e^{e^4}\right) = e^4$

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

Here are 8 related questions to expand on the topic:

1. How would you express $\log_2 32$ in its simplest form?
2. Can you simplify $\ln(e^7)$?
3. What is the value of $\log_3 \frac{27}{81}$?
4. How do you solve $\log_7 49$?
5. What is the result of $\ln(\frac{1}{e^3})$?
6. How do you simplify $\log_{10} 1000$?
7. Can you calculate $\log_8 64$?
8. How would you solve $\log_2 0.5$?

Tip: Always remember that logarithms and exponentials are inverse operations. This is key to simplifying many expressions.