Evaluating Logarithms worksheets - Free Printable

Let's evaluate each logarithmic expression step by step. The key idea is to rewrite the argument (the number inside the log) as a power of the base, then use the definition:

> If $ \log_b(a) = x $, then $ b^x = a $

---

1) $ \log_{11} 121 $

We know $ 11^2 = 121 $, so:
$$
\log_{11} 121 = 2
$$

---

2) $ \log_9 81 $

$ 9^2 = 81 $, so:
$$
\log_9 81 = 2
$$

---

3) $ \log_7 49 $

$ 7^2 = 49 $, so:
$$
\log_7 49 = 2
$$

---

4) $ \log_2 128 $

$ 128 = 2^7 $, so:
$$
\log_2 128 = 7
$$

---

5) $ \log_{12} 144 $

$ 12^2 = 144 $, so:
$$
\log_{12} 144 = 2
$$

---

6) $ \log_8 2 $

We want $ 8^x = 2 $. But $ 8 = 2^3 $, so:
$$
(2^3)^x = 2 \Rightarrow 2^{3x} = 2^1 \Rightarrow 3x = 1 \Rightarrow x = \frac{1}{3}
$$
So:
$$
\log_8 2 = \frac{1}{3}
$$

---

7) $ \log_5 125 $

$ 5^3 = 125 $, so:
$$
\log_5 125 = 3
$$

---

8) $ \log_3 81 $

$ 3^4 = 81 $, so:
$$
\log_3 81 = 4
$$

---

9) $ \log_{32} 2 $

$ 32 = 2^5 $, so $ 32^x = 2 \Rightarrow (2^5)^x = 2 \Rightarrow 2^{5x} = 2^1 \Rightarrow 5x = 1 \Rightarrow x = \frac{1}{5} $
$$
\log_{32} 2 = \frac{1}{5}
$$

---

10) $ \log_5 25 $

$ 5^2 = 25 $, so:
$$
\log_5 25 = 2
$$

---

11) $ \log_4 2 $

$ 4 = 2^2 $, so $ 4^x = 2 \Rightarrow (2^2)^x = 2 \Rightarrow 2^{2x} = 2^1 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2} $
$$
\log_4 2 = \frac{1}{2}
$$

---

12) $ \log_{10} 100 $

$ 10^2 = 100 $, so:
$$
\log_{10} 100 = 2
$$

---

13) $ \log_{10} 1000 $

$ 10^3 = 1000 $, so:
$$
\log_{10} 1000 = 3
$$

---

14) $ \log_{\frac{1}{2}} 128 $

This one is trickier.

We want $ \left(\frac{1}{2}\right)^x = 128 $

But $ \frac{1}{2} = 2^{-1} $, and $ 128 = 2^7 $, so:
$$
(2^{-1})^x = 2^7 \Rightarrow 2^{-x} = 2^7 \Rightarrow -x = 7 \Rightarrow x = -7
$$
So:
$$
\log_{\frac{1}{2}} 128 = -7
$$

---

15) $ \log_4 64 $

$ 64 = 4^x $? Note: $ 4 = 2^2 $, $ 64 = 2^6 $

So $ (2^2)^x = 2^6 \Rightarrow 2^{2x} = 2^6 \Rightarrow 2x = 6 \Rightarrow x = 3 $

Alternatively: $ 4^3 = (2^2)^3 = 2^6 = 64 $

So:
$$
\log_4 64 = 3
$$

---

16) $ \log_3 27 $

$ 3^3 = 27 $, so:
$$
\log_3 27 = 3
$$

---

17) $ \log_{27} 3 $

$ 27 = 3^3 $, so $ 27^x = 3 \Rightarrow (3^3)^x = 3 \Rightarrow 3^{3x} = 3^1 \Rightarrow 3x = 1 \Rightarrow x = \frac{1}{3} $
$$
\log_{27} 3 = \frac{1}{3}
$$

---

18) $ \log_{13} 169 $

$ 13^2 = 169 $, so:
$$
\log_{13} 169 = 2
$$

---

19) $ \log_5 625 $

$ 5^4 = 625 $, so:
$$
\log_5 625 = 4
$$

---

20) $ \log_6 216 $

$ 6^3 = 216 $, so:
$$
\log_6 216 = 3
$$

---

21) $ \log_{32} 128 $

$ 32 = 2^5 $, $ 128 = 2^7 $

So $ (2^5)^x = 2^7 \Rightarrow 2^{5x} = 2^7 \Rightarrow 5x = 7 \Rightarrow x = \frac{7}{5} $
$$
\log_{32} 128 = \frac{7}{5}
$$

---

✔ Final Answers:

| # | Answer |
|---|--------|
| 1) | 2 |
| 2) | 2 |
| 3) | 2 |
| 4) | 7 |
| 5) | 2 |
| 6) | $ \frac{1}{3} $ |
| 7) | 3 |
| 8) | 4 |
| 9) | $ \frac{1}{5} $ |
| 10) | 2 |
| 11) | $ \frac{1}{2} $ |
| 12) | 2 |
| 13) | 3 |
| 14) | -7 |
| 15) | 3 |
| 16) | 3 |
| 17) | $ \frac{1}{3} $ |
| 18) | 2 |
| 19) | 4 |
| 20) | 3 |
| 21) | $ \frac{7}{5} $ |

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🔍 Summary Tips:

- When the base and argument are powers of the same number, rewrite both in terms of that base.
- Use the identity: $ \log_b(a) = x $ iff $ b^x = a $
- For fractional bases like $ \frac{1}{2} $, remember $ \left(\frac{1}{2}\right)^x = 2^{-x} $
- Always check if the argument is a perfect power of the base.

Let me know if you'd like this formatted as a printable worksheet or with explanations for each!