Logarithmic expressions and equations in a structured format.

Let's solve each logarithmic expression step by step, using logarithmic identities and properties.

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Logarithmic Identities Used:

1. $\log_b b = 1$
2. $\log_b (a^n) = n \log_b a$
3. $\log_b (a \cdot c) = \log_b a + \log_b c$
4. $\log_b \left(\frac{a}{c}\right) = \log_b a - \log_b c$
5. $\log_{b^m} a = \frac{1}{m} \log_b a$
6. $\log_b a = \frac{\log_c a}{\log_c b}$ (Change of base)
7. $\log_b 1 = 0$
8. $\ln e^x = x$, $\log_{10} 10^x = x$
9. $\log_b \left(\frac{1}{a}\right) = -\log_b a$

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Now let’s go through each part:

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a) $\log 1$
$\log 1 = 0$ (since $10^0 = 1$)
✔ Answer: 0

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b) $\log_3 27$
$27 = 3^3$, so $\log_3 27 = \log_3 (3^3) = 3$
✔ Answer: 3

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c) $\log_4 16$
$16 = 4^2$, so $\log_4 16 = \log_4 (4^2) = 2$
✔ Answer: 2

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d) $\log_2 16$
$16 = 2^4$, so $\log_2 16 = 4$
✔ Answer: 4

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e) $\log_4 2$
Note: $2 = 4^{1/2}$, so $\log_4 2 = \log_4 (4^{1/2}) = \frac{1}{2}$
✔ Answer: $\frac{1}{2}$

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f) $\log_{27} 3$
Let $x = \log_{27} 3$. Then $27^x = 3$. But $27 = 3^3$, so $(3^3)^x = 3^{3x} = 3^1$ → $3x = 1$ → $x = \frac{1}{3}$
✔ Answer: $\frac{1}{3}$

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g) $\log 10,000$
$10,000 = 10^4$, so $\log_{10} 10^4 = 4$
✔ Answer: 4

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h) $\log_3 9^2$
First, $9^2 = (3^2)^2 = 3^4$, so $\log_3 9^2 = \log_3 (3^4) = 4$
✔ Answer: 4

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i) $\log_{\frac{1}{9}} 3$
Let $x = \log_{\frac{1}{9}} 3$. Then $\left(\frac{1}{9}\right)^x = 3$.
But $\frac{1}{9} = 9^{-1} = (3^2)^{-1} = 3^{-2}$, so:
$(3^{-2})^x = 3^{-2x} = 3^1$ → $-2x = 1$ → $x = -\frac{1}{2}$
✔ Answer: $-\frac{1}{2}$

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j) $\log_{\frac{1}{3}} \frac{1}{9}$
$\frac{1}{9} = \left(\frac{1}{3}\right)^2$, so $\log_{\frac{1}{3}} \left(\left(\frac{1}{3}\right)^2\right) = 2$
✔ Answer: 2

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k) $\ln e^2$
$\ln e^2 = 2 \ln e = 2 \cdot 1 = 2$
✔ Answer: 2

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l) $\log_{\frac{1}{5}} 125$
$125 = 5^3$, and $\frac{1}{5} = 5^{-1}$, so:
Let $x = \log_{5^{-1}} (5^3)$ → $(5^{-1})^x = 5^3$ → $5^{-x} = 5^3$ → $-x = 3$ → $x = -3$
✔ Answer: $-3$

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m) $\log_{216} 6^{-1}$
$216 = 6^3$, so $\log_{6^3} (6^{-1}) = \frac{\log_6 (6^{-1})}{\log_6 (6^3)} = \frac{-1}{3} = -\frac{1}{3}$
Alternatively: Let $x = \log_{6^3} (6^{-1})$, then $(6^3)^x = 6^{-1}$ → $6^{3x} = 6^{-1}$ → $3x = -1$ → $x = -\frac{1}{3}$
✔ Answer: $-\frac{1}{3}$

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n) $\log 10^{10}$
$\log_{10} (10^{10}) = 10$
✔ Answer: 10

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o) $\log_8 64$
$64 = 8^?$. Note: $8 = 2^3$, $64 = 2^6$, so $64 = (2^3)^{6/3} = 8^2$, so $\log_8 64 = 2$
✔ Answer: 2

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p) $\log_{\frac{1}{9}} 81$
$81 = 9^2$, and $\frac{1}{9} = 9^{-1}$, so:
$\log_{9^{-1}} (9^2) = \frac{\log_9 (9^2)}{\log_9 (9^{-1})} = \frac{2}{-1} = -2$
Or: $(9^{-1})^x = 9^2$ → $9^{-x} = 9^2$ → $-x = 2$ → $x = -2$
✔ Answer: $-2$

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q) $\log_{21} 21$
Any log of the base is 1 → $\log_{21} 21 = 1$
✔ Answer: 1

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r) $\log_{\sqrt{51}} 51^2$
Let’s simplify: $\sqrt{51} = 51^{1/2}$, so:
$\log_{51^{1/2}} (51^2) = \frac{\log_{51} (51^2)}{\log_{51} (51^{1/2})} = \frac{2}{1/2} = 4$
✔ Answer: 4

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s) $\log_{128} \frac{1}{4}$
$128 = 2^7$, $4 = 2^2$, so $\frac{1}{4} = 2^{-2}$
So: $\log_{2^7} (2^{-2}) = \frac{\log_2 (2^{-2})}{\log_2 (2^7)} = \frac{-2}{7} = -\frac{2}{7}$
✔ Answer: $-\frac{2}{7}$

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t) $\log_{\frac{1}{8}} \sqrt{128}$
First, $\sqrt{128} = 128^{1/2} = (2^7)^{1/2} = 2^{7/2}$
Base: $\frac{1}{8} = 8^{-1} = (2^3)^{-1} = 2^{-3}$
So: $\log_{2^{-3}} (2^{7/2}) = \frac{\log_2 (2^{7/2})}{\log_2 (2^{-3})} = \frac{7/2}{-3} = -\frac{7}{6}$
✔ Answer: $-\frac{7}{6}$

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u) $\log_3 18 + \log_3 \frac{3}{2}$
Use: $\log_3 (18 \cdot \frac{3}{2}) = \log_3 (27) = \log_3 (3^3) = 3$
✔ Answer: 3

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v) $\log e^2 + \log e^{-2}$
Assuming base 10: $\log e^2 = 2 \log e$, $\log e^{-2} = -2 \log e$
So total: $2 \log e - 2 \log e = 0$
✔ Answer: 0

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w) $\ln e^2 + \ln e^3$
$\ln e^2 = 2$, $\ln e^3 = 3$, sum = $2 + 3 = 5$
✔ Answer: 5

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x) $\log 20 + \log 50$
= $\log (20 \cdot 50) = \log (1000) = \log (10^3) = 3$
✔ Answer: 3

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y) $\log 500 - \log 5$
= $\log \left(\frac{500}{5}\right) = \log (100) = \log (10^2) = 2$
✔ Answer: 2

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z) $\log 200 + \log 5 - \log 100$
= $\log \left(\frac{200 \cdot 5}{100}\right) = \log \left(\frac{1000}{100}\right) = \log (10) = 1$
✔ Answer: 1

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Z) $\log_5 20 + \log_5 \frac{125}{4} - \log_5 5^{-1}$
Use log rules:
= $\log_5 \left(20 \cdot \frac{125}{4} \div 5^{-1}\right)$
First, $20 \cdot \frac{125}{4} = 5 \cdot 125 = 625$
Then divide by $5^{-1}$: $625 \div 5^{-1} = 625 \cdot 5 = 3125$
So: $\log_5 3125$
Now $3125 = 5^5$, since $5^5 = 3125$
Thus: $\log_5 5^5 = 5$
✔ Answer: 5

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✔ Final Answers:

| Letter | Answer |
|--------|--------|
| a) | 0 |
| b) | 3 |
| c) | 2 |
| d) | 4 |
| e) | ½ |
| f) | ⅓ |
| g) | 4 |
| h) | 4 |
| i) | -½ |
| j) | 2 |
| k) | 2 |
| l) | -3 |
| m) | -⅓ |
| n) | 10 |
| o) | 2 |
| p) | -2 |
| q) | 1 |
| r) | 4 |
| s) | -2/7 |
| t) | -7/6 |
| u) | 3 |
| v) | 0 |
| w) | 5 |
| x) | 3 |
| y) | 2 |
| z) | 1 |
| Z) | 5 |

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