Evaluating Logarithms Worksheet: Complete with ease | airSlate SignNow - Free Printable

Let's solve the Practice Worksheet: Evaluating Logarithms step by step. I'll go through each section, rewrite equations in the required form, and evaluate the logarithms.

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Section 1: Rewrite the equation in exponential form

We use the definition:
> If $ \log_b a = c $, then $ b^c = a $

1) $ \log_2 64 = 6 $
→ $ 2^6 = 64 $

2) $ \log_5 125 = 3 $
→ $ 5^3 = 125 $

3) $ \log_{10} \frac{1}{100} = -2 $
→ $ 10^{-2} = \frac{1}{100} $

4) $ \log_8 16 = x $
→ $ 8^x = 16 $

5) $ \log_{1/2} 4 = \frac{1}{2} $
→ $ \left(\frac{1}{2}\right)^{1/2} = 4 $? Wait — this is incorrect! Let's check:

Wait: $ \left(\frac{1}{2}\right)^{1/2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \neq 4 $. So something’s wrong.

But the log says $ \log_{1/2} 4 = \frac{1}{2} $. That would mean $ (1/2)^{1/2} = 4 $, which is false.

Let’s double-check:
We want $ \left(\frac{1}{2}\right)^x = 4 $
$ \Rightarrow \left(2^{-1}\right)^x = 2^2 $
$ \Rightarrow 2^{-x} = 2^2 $ → $ -x = 2 $ → $ x = -2 $

So $ \log_{1/2} 4 = -2 $, not $ \frac{1}{2} $. But the problem says $ \log_{1/2} 4 = \frac{1}{2} $ — that must be a typo.

Wait — let's look again:
It says:
5) $ \log_{1/2} 4 = \frac{1}{2} $

This is false. But maybe it's meant to be $ \log_{1/2} \frac{1}{4} = 2 $? Or perhaps we're just rewriting as given.

But since the instruction is to "rewrite the equation in exponential form", even if it's incorrect, we follow it.

So:
5) $ \log_{1/2} 4 = \frac{1}{2} $ → $ \left(\frac{1}{2}\right)^{1/2} = 4 $
But this is not true — so likely a mistake. But for now, we write what it implies.

But wait — perhaps it's a typo and should be $ \log_{1/2} \frac{1}{4} = 2 $? Because $ (1/2)^2 = 1/4 $

Alternatively, maybe it's $ \log_{1/2} \frac{1}{4} = 2 $, but here it says $ \log_{1/2} 4 = \frac{1}{2} $, which is wrong.

But we are to rewrite it in exponential form, so regardless of truth, we do:

5) $ \left(\frac{1}{2}\right)^{1/2} = 4 $ → False, but that’s what it implies.

But let's assume it's correct for the sake of exercise? No — better to point out.

Actually, maybe it's $ \log_{1/2} \frac{1}{4} = 2 $? Let me recheck.

Wait — perhaps it's $ \log_{1/2} 4 = -2 $? Then $ (1/2)^{-2} = 4 $, yes.

So likely a typo: probably meant $ \log_{1/2} 4 = -2 $

But since the worksheet says $ = \frac{1}{2} $, I’ll proceed with what’s written.

So:

5) $ \log_{1/2} 4 = \frac{1}{2} $ → $ \left(\frac{1}{2}\right)^{1/2} = 4 $ — Incorrect, but per instruction, we write it.

But let's move on — perhaps it’s intended to be correct.

Wait — no: $ (1/2)^{1/2} = \sqrt{1/2} \approx 0.707 $, not 4.

So clearly wrong.

Maybe it's $ \log_4 (1/2) = -\frac{1}{2} $? Not matching.

Alternatively, maybe it's $ \log_{1/2} (1/4) = 2 $? Yes: $ (1/2)^2 = 1/4 $

But the problem says $ \log_{1/2} 4 = \frac{1}{2} $ — that’s incorrect.

I suspect a typo. But since we’re just rewriting, we do:

Answer: $ \left(\frac{1}{2}\right)^{1/2} = 4 $ — though false.

But let's skip ahead and come back.

6) $ \log_2 8 = -2 $? No — $ \log_2 8 = 3 $, because $ 2^3 = 8 $

Wait — problem says: 6) $ \log_2 8 = -2 $? That can’t be.

Wait — no: the problem says:

6) $ \log_2 8 = -2 $? Let's look:

No — actually, the list is:

1) $ \log_2 64 = 6 $ → $ 2^6 = 64 $ ✔
2) $ \log_5 125 = 3 $ → $ 5^3 = 125 $ ✔
3) $ \log_{10} \frac{1}{100} = -2 $ → $ 10^{-2} = \frac{1}{100} $ ✔
4) $ \log_8 16 = x $ → $ 8^x = 16 $ ✔
5) $ \log_{1/2} 4 = \frac{1}{2} $ → $ (1/2)^{1/2} = 4 $ ✘
6) $ \log_2 8 = -2 $ → $ 2^{-2} = 8 $? $ 2^{-2} = 1/4 $, not 8 → ✘

So both 5 and 6 are incorrect.

But perhaps they are meant to be rewritten as-is, even if false?

Possibly — or maybe typos.

But let’s assume they are correct and proceed.

So:

5) $ \log_{1/2} 4 = \frac{1}{2} $ → $ \left(\frac{1}{2}\right)^{1/2} = 4 $
6) $ \log_2 8 = -2 $ → $ 2^{-2} = 8 $ → False

But again, we’re just rewriting.

Wait — maybe it's $ \log_2 \frac{1}{8} = -3 $? But not matching.

Alternatively, perhaps the problem has a typo.

But let's move on — maybe later we’ll see.

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Section 2: Rewrite in logarithmic form

Use: $ b^c = a $ → $ \log_b a = c $

7) $ 13^2 = 169 $ → $ \log_{13} 169 = 2 $

8) $ 9^{0.5} = 27 $? Wait: $ 9^{0.5} = \sqrt{9} = 3 $, not 27 → So this is false.

Wait: $ 9^{0.5} = 3 $, not 27.

But the problem says: 8) $ 9^{0.5} = 27 $

That’s incorrect.

$ 9^{0.5} = 3 $, $ 3^3 = 27 $, so $ 9^{1.5} = (3^2)^{1.5} = 3^3 = 27 $

So $ 9^{1.5} = 27 $, not $ 9^{0.5} = 27 $

So likely typo: should be $ 9^{1.5} = 27 $

But as written: $ 9^{0.5} = 27 $ → $ \log_9 27 = 0.5 $? But $ \log_9 27 = \frac{\log 27}{\log 9} = \frac{3\log 3}{2\log 3} = \frac{3}{2} $

So $ \log_9 27 = 1.5 $, not 0.5

So again, error.

But we’ll rewrite as per instruction:

8) $ 9^{0.5} = 27 $ → $ \log_9 27 = 0.5 $ — False

9) $ 4^x = \frac{1}{16} $ → $ \log_4 \left(\frac{1}{16}\right) = x $

10) $ 10^3 = 0.001 $? $ 10^3 = 1000 $, not 0.001 → $ 10^{-3} = 0.001 $

So likely typo: should be $ 10^{-3} = 0.001 $

But as written: $ 10^3 = 0.001 $ → $ \log_{10} 0.001 = 3 $ — false

11) $ 6^d = 8 $ → $ \log_6 8 = d $

12) $ 9^x = \frac{1}{27} $ → $ \log_9 \left(\frac{1}{27}\right) = x $

13) $ 12^a = 144 $ → $ \log_{12} 144 = a $

14) $ \left(\frac{1}{3}\right)^x = \frac{1}{243} $ → $ \log_{1/3} \left(\frac{1}{243}\right) = x $

Now, note: $ \frac{1}{243} = 3^{-5} $, and $ \frac{1}{3} = 3^{-1} $, so $ (3^{-1})^x = 3^{-x} = 3^{-5} $ → $ x = 5 $

So $ \log_{1/3}(1/243) = 5 $

But we just need to rewrite.

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Section 3: Evaluate the logarithm without using a calculator

We’ll use properties of logs:

#### 15) $ \log_3 81 $

$ 81 = 3^4 $ → $ \log_3 3^4 = 4 $

✔ Answer: 4

#### 16) $ \log_2 2 $ → $ \log_2 2 = 1 $

✔ Answer: 1

#### 17) $ \log_7 49 $

$ 49 = 7^2 $ → $ \log_7 7^2 = 2 $

✔ Answer: 2

#### 18) $ \log_9 1 $

Any log of 1 is 0 → $ \log_9 1 = 0 $

✔ Answer: 0

#### 19) $ \ln e^4 $

Natural log: $ \ln(e^4) = 4 $

✔ Answer: 4

#### 20) $ \log_8 4 $

We know $ 8 = 2^3 $, $ 4 = 2^2 $

Let $ x = \log_8 4 $ → $ 8^x = 4 $

$ (2^3)^x = 2^2 $ → $ 2^{3x} = 2^2 $ → $ 3x = 2 $ → $ x = \frac{2}{3} $

✔ Answer: $ \frac{2}{3} $

#### 21) $ \log_2 \left(\frac{1}{8}\right) $

$ \frac{1}{8} = 2^{-3} $ → $ \log_2 2^{-3} = -3 $

✔ Answer: -3

#### 22) $ \log 10000 $

$ 10000 = 10^4 $ → $ \log_{10} 10^4 = 4 $

✔ Answer: 4

#### 23) $ \log_3 1/27 $

$ 1/27 = 3^{-3} $ → $ \log_3 3^{-3} = -3 $

✔ Answer: -3

#### 24) $ \log_4 2 $

$ 4 = 2^2 $, $ 2 = 4^{1/2} $ → $ \log_4 2 = \frac{1}{2} $

Or: $ x = \log_4 2 $ → $ 4^x = 2 $ → $ (2^2)^x = 2 $ → $ 2^{2x} = 2^1 $ → $ 2x = 1 $ → $ x = \frac{1}{2} $

✔ Answer: $ \frac{1}{2} $

#### 25) $ \log_2 325 $

$ 325 $ — not a power of 2. But $ 2^8 = 256 $, $ 2^9 = 512 $ → between 8 and 9

But we need exact value? It's not a nice number.

Wait — is it $ \log_2 32 $ or $ \log_2 256 $? 325 is not a power of 2.

But 325 = ? Let's see: $ 2^8 = 256 $, $ 2^9 = 512 $, so $ \log_2 325 $ is between 8 and 9.

But the problem says “without using a calculator” — so unless it's a trick, maybe it's a typo?

Wait — could it be $ \log_2 32 $ or $ \log_2 256 $? 325 isn't a power of 2.

Wait — maybe it's $ \log_2 32 $ → $ 32 = 2^5 $ → $ \log_2 32 = 5 $

But it says 325.

Perhaps $ \log_2 32 $ was meant.

But as written: $ \log_2 325 $

But 325 = 25 × 13 = 5² × 13 → not a power of 2.

So cannot simplify exactly.

But maybe it's $ \log_2 32 $ or $ \log_2 256 $? Or $ \log_2 256 $?

Wait — maybe it's $ \log_2 256 $ → $ 2^8 = 256 $ → 8

But it says 325.

Wait — could it be $ \log_2 32 $?

But it says 325.

Wait — maybe it's $ \log_2 32 $, typo.

Alternatively, perhaps $ \log_2 32 $ is meant.

But let's assume it's $ \log_2 325 $ — then we leave it as is? But the problem asks to evaluate.

Wait — maybe it's $ \log_2 32 $, and typo.

But looking at others: 26) $ \log_3 243 $ — $ 243 = 3^5 $ → $ \log_3 243 = 5 $

So maybe 25) is $ \log_2 32 $ → $ 32 = 2^5 $ → 5

Yes — likely typo: 325 → 32

So assume $ \log_2 32 = 5 $

✔ Answer: 5

#### 26) $ \log_3 243 $

$ 243 = 3^5 $ → $ \log_3 3^5 = 5 $

✔ Answer: 5

#### 27) $ \log_4 64 $

$ 64 = 4^x $? $ 4 = 2^2 $, $ 64 = 2^6 $, so $ 4^x = (2^2)^x = 2^{2x} = 2^6 $ → $ 2x = 6 $ → $ x = 3 $

Or $ 64 = 4^3 $? $ 4^3 = 64 $ → yes!

So $ \log_4 64 = 3 $

✔ Answer: 3

#### 28) $ \log_4 4 $ → $ \log_4 4 = 1 $

✔ Answer: 1

#### 29) $ \log_2 \left(\frac{1}{2}\right) $

$ \frac{1}{2} = 2^{-1} $ → $ \log_2 2^{-1} = -1 $

✔ Answer: -1

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Summary of Answers

#### Section 1: Exponential Form
1) $ 2^6 = 64 $
2) $ 5^3 = 125 $
3) $ 10^{-2} = \frac{1}{100} $
4) $ 8^x = 16 $
5) $ \left(\frac{1}{2}\right)^{1/2} = 4 $ ← Incorrect, but as per problem
6) $ 2^{-2} = 8 $ ← Incorrect

Note: Problems 5 and 6 appear to have typos.

#### Section 2: Logarithmic Form
7) $ \log_{13} 169 = 2 $
8) $ \log_9 27 = 0.5 $ ← Incorrect (should be 1.5)
9) $ \log_4 \left(\frac{1}{16}\right) = x $
10) $ \log_{10} 0.001 = 3 $ ← Incorrect (should be -3)
11) $ \log_6 8 = d $
12) $ \log_9 \left(\frac{1}{27}\right) = x $
13) $ \log_{12} 144 = a $
14) $ \log_{1/3} \left(\frac{1}{243}\right) = x $

#### Section 3: Evaluate
15) 4
16) 1
17) 2
18) 0
19) 4
20) $ \frac{2}{3} $
21) -3
22) 4
23) -3
24) $ \frac{1}{2} $
25) Assuming typo: $ \log_2 32 = 5 $
26) 5
27) 3
28) 1
29) -1

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Final Answer Key (Corrected)

| Problem | Answer |
|--------|--------|
| 1 | $ 2^6 = 64 $ |
| 2 | $ 5^3 = 125 $ |
| 3 | $ 10^{-2} = \frac{1}{100} $ |
| 4 | $ 8^x = 16 $ |
| 5 | $ \left(\frac{1}{2}\right)^{1/2} = 4 $ (but incorrect; should be $ \log_{1/2} 4 = -2 $) |
| 6 | $ 2^{-2} = 8 $ (incorrect; should be $ \log_2 8 = 3 $) |
| 7 | $ \log_{13} 169 = 2 $ |
| 8 | $ \log_9 27 = 0.5 $ (incorrect; should be $ \log_9 27 = 1.5 $) |
| 9 | $ \log_4 \left(\frac{1}{16}\right) = x $ |
|10 | $ \log_{10} 0.001 = 3 $ (incorrect; should be $ \log_{10} 0.001 = -3 $) |
|11 | $ \log_6 8 = d $ |
|12 | $ \log_9 \left(\frac{1}{27}\right) = x $ |
|13 | $ \log_{12} 144 = a $ |
|14 | $ \log_{1/3} \left(\frac{1}{243}\right) = x $ |
|15 | 4 |
|16 | 1 |
|17 | 2 |
|18 | 0 |
|19 | 4 |
|20 | $ \frac{2}{3} $ |
|21 | -3 |
|22 | 4 |
|23 | -3 |
|24 | $ \frac{1}{2} $ |
|25 | 5 (assuming $ \log_2 32 $ instead of 325) |
|26 | 5 |
|27 | 3 |
|28 | 1 |
|29 | -1 |

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Notes:

- Several problems (5, 6, 8, 10) contain errors in the given equations.
- Likely typos in the worksheet.
- For evaluation, most answers are exact and simple.

If you'd like, I can provide a corrected version of the worksheet.