Solved Intro to Logarithms Worksheet Properties of Simple | Chegg.com - Free Printable

Let's evaluate each of the logarithmic expressions without using a calculator, and we'll use exponential form where helpful. We’ll go step by step.

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9) $ \log_3 81 = ? $

We know:
$$
81 = 3^4
$$
So:
$$
\log_3 81 = \log_3 (3^4) = 4
$$
✔ Answer: 4

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10) $ \log_4 \frac{1}{2} = ? $

We want to find $ x $ such that:
$$
4^x = \frac{1}{2}
$$

Note: $ 4 = 2^2 $, so:
$$
(2^2)^x = 2^{2x} = \frac{1}{2} = 2^{-1}
\Rightarrow 2x = -1 \Rightarrow x = -\frac{1}{2}
$$
✔ Answer: $ -\frac{1}{2} $

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11) $ \log_{12} 144 = ? $

We know:
$$
144 = 12^2
\Rightarrow \log_{12} 144 = \log_{12} (12^2) = 2
$$
✔ Answer: 2

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12) $ \log_6 \frac{1}{36} = ? $

$$
\frac{1}{36} = 6^{-2}, \quad \text{since } 6^2 = 36
\Rightarrow \log_6 \left(6^{-2}\right) = -2
$$
✔ Answer: -2

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13) $ \log_{\frac{1}{3}} \frac{9}{4} = ? $

We want $ x $ such that:
$$
\left(\frac{1}{3}\right)^x = \frac{9}{4}
$$

But $ \frac{1}{3} = 3^{-1} $, so:
$$
(3^{-1})^x = 3^{-x} = \frac{9}{4}
$$

But $ \frac{9}{4} = \left(\frac{3}{2}\right)^2 $. This is not a power of 3, so maybe rewrite differently?

Wait — let’s try expressing both sides in terms of base 3 and 2:

We have:
$$
\left(\frac{1}{3}\right)^x = \frac{9}{4} = \frac{3^2}{2^2}
$$

But left side is $ 3^{-x} $, which has no factor of 2 → contradiction unless we approach differently.

Actually, this suggests it might not be an integer. Let's solve:
$$
\left(\frac{1}{3}\right)^x = \frac{9}{4} \Rightarrow 3^{-x} = \frac{9}{4}
\Rightarrow -x = \log_3 \left(\frac{9}{4}\right) = \log_3 9 - \log_3 4 = 2 - \log_3 4
$$

But we are told to avoid calculators and possibly expect exact values.

Wait — perhaps there's a mistake. Let's double-check.

Is $ \left(\frac{1}{3}\right)^x = \frac{9}{4} $? Then:
$$
\left(\frac{1}{3}\right)^x = \frac{9}{4} \Rightarrow \left(3^{-1}\right)^x = 3^{-x} = \frac{9}{4}
\Rightarrow 3^{-x} = \frac{9}{4}
\Rightarrow -x = \log_3 \left(\frac{9}{4}\right)
\Rightarrow x = -\log_3 \left(\frac{9}{4}\right)
$$

But this doesn't simplify nicely. Wait — maybe it's supposed to be $ \log_{\frac{1}{3}} \frac{1}{9} $? But it's $ \frac{9}{4} $.

Alternatively, maybe it's a typo? But let's proceed.

Wait — actually, perhaps express $ \frac{9}{4} $ as $ \left(\frac{3}{2}\right)^2 $. But base is $ \frac{1}{3} $, so unless we can write $ \left(\frac{1}{3}\right)^x = \left(\frac{3}{2}\right)^2 $, but that’s not helpful.

Alternatively, use change of base:
$$
\log_{\frac{1}{3}} \frac{9}{4} = \frac{\log \frac{9}{4}}{\log \frac{1}{3}} = \frac{\log 9 - \log 4}{\log 1 - \log 3} = \frac{2\log 3 - 2\log 2}{-\log 3} = \frac{2(\log 3 - \log 2)}{-\log 3}
= -2\left(1 - \frac{\log 2}{\log 3}\right)
$$

This is messy — probably not intended.

Wait — maybe the problem is $ \log_{\frac{1}{3}} \frac{1}{9} $? Then:
$$
\left(\frac{1}{3}\right)^x = \frac{1}{9} = \left(\frac{1}{3}\right)^2 \Rightarrow x = 2
$$

But it's written as $ \frac{9}{4} $, so likely not.

Wait — another idea: could it be $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) $? No, it says $ \frac{9}{4} $.

Perhaps it's $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) $? But it's clearly $ \frac{9}{4} $.

Let’s assume it's correct. Then:
$$
\left(\frac{1}{3}\right)^x = \frac{9}{4}
\Rightarrow 3^{-x} = \frac{9}{4} = \frac{3^2}{2^2}
\Rightarrow 3^{-x} = 3^2 \cdot 2^{-2}
\Rightarrow \text{No solution with only powers of 3}
$$

So unless we allow logs, it doesn’t simplify.

Wait — perhaps it's $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) $? But no.

Wait — maybe the base is $ \frac{1}{3} $, and argument is $ \frac{1}{9} $? But it's $ \frac{9}{4} $.

Hmm. Perhaps it's a trick question?

Alternatively, let's suppose it's $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) $. Then:
$$
\left(\frac{1}{3}\right)^x = \frac{1}{9} = \left(\frac{1}{3}\right)^2 \Rightarrow x = 2
$$

But the problem says $ \frac{9}{4} $, so I think it might be a typo or requires different approach.

Wait — perhaps we can write:
$$
\left(\frac{1}{3}\right)^x = \frac{9}{4} \Rightarrow 3^{-x} = \frac{9}{4} \Rightarrow -x = \log_3(9/4) = \log_3 9 - \log_3 4 = 2 - \log_3 4
\Rightarrow x = -2 + \log_3 4
$$

But again, not nice.

Wait — perhaps the problem is $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) $? That would make sense.

But since it says $ \frac{9}{4} $, and we can't simplify it nicely, maybe skip for now.

Wait — check if $ \left(\frac{1}{3}\right)^x = \frac{9}{4} $ has a solution.

Try $ x = -1 $: $ \left(\frac{1}{3}\right)^{-1} = 3 $

$ x = -2 $: $ 9 $

$ x = -1.5 $: $ \left(\frac{1}{3}\right)^{-1.5} = 3^{1.5} = 3\sqrt{3} \approx 5.2 $, too big

But $ \frac{9}{4} = 2.25 $, so between $ x = -1 $ and $ x = -2 $, but not rational.

So likely, either typo or not meant to be simplified.

Wait — perhaps it's $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) $? Then answer is 2.

But since it's written as $ \frac{9}{4} $, I think we must accept it's messy.

Alternatively, maybe it's $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) $, and $ \frac{9}{4} $ is a typo.

But let's move on and come back.

Wait — perhaps it's $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) $? But it's $ \frac{9}{4} $.

I think there may be a typo. Let's assume it's $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) = 2 $, but we'll keep as is.

Wait — another idea: $ \frac{9}{4} = \left(\frac{3}{2}\right)^2 $, and $ \frac{1}{3} = 3^{-1} $, but no direct relation.

So perhaps this one is not simplifiable without logs.

But let's look at others.

Maybe skip for now.

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14) $ \log_{0.25} 4 = ? $

Note: $ 0.25 = \frac{1}{4} = 4^{-1} $

So:
$$
\log_{4^{-1}} 4 = \frac{\log_4 4}{\log_4 (4^{-1})} = \frac{1}{-1} = -1
$$

Alternatively:
Let $ x = \log_{0.25} 4 $, then:
$$
(0.25)^x = 4 \Rightarrow \left(\frac{1}{4}\right)^x = 4 \Rightarrow 4^{-x} = 4^1 \Rightarrow -x = 1 \Rightarrow x = -1
$$
✔ Answer: -1

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15) $ \log_3 (-3) = ? $

Logarithm of a negative number is undefined in real numbers.

✔ Answer: undefined

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16) $ \log_4 4 = ? $

By definition: $ \log_b b = 1 $

✔ Answer: 1

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17) $ \log_{64} \frac{1}{27} = ? $

Note: $ 64 = 2^6 $, $ 27 = 3^3 $

But base and argument are different primes — not easy.

Wait — $ 64 = 4^3 $, $ 27 = 3^3 $, still not helpful.

Wait — $ 64 = 8^2 $, $ 27 = 3^3 $

Try writing both in exponential form.

Let $ x = \log_{64} \frac{1}{27} $

Then:
$$
64^x = \frac{1}{27}
\Rightarrow (2^6)^x = 2^{6x} = 3^{-3}
$$

Left side is power of 2, right side power of 3 — no solution unless we use logs.

But wait — perhaps express 64 and 27 in same base?

No.

Wait — $ 64 = 2^6 $, $ 27 = 3^3 $, so $ \frac{1}{27} = 3^{-3} $

But $ 64^x = (2^6)^x = 2^{6x} $, which cannot equal $ 3^{-3} $ unless complex numbers.

So unless we use change of base:

$$
\log_{64} \frac{1}{27} = \frac{\log \frac{1}{27}}{\log 64} = \frac{-\log 27}{\log 64} = \frac{-\log(3^3)}{\log(2^6)} = \frac{-3\log 3}{6\log 2} = -\frac{1}{2} \cdot \frac{\log 3}{\log 2} = -\frac{1}{2} \log_2 3
$$

Not a nice number. So perhaps not intended.

Wait — is $ 64 = 4^3 $, $ 27 = 3^3 $, so $ \frac{1}{27} = 3^{-3} $, $ 64 = 4^3 $, but 4 and 3 unrelated.

Unless we notice: $ 64 = 2^6 $, $ 27 = 3^3 $, so $ \frac{1}{27} = 3^{-3} $, but no common base.

So likely, this one is not simplifiable to a rational number.

But wait — maybe it's $ \log_{64} \frac{1}{8} $? $ 8 = 2^3 $, $ 64 = 2^6 $, so $ \log_{64} \frac{1}{8} = \log_{2^6} (2^{-3}) = \frac{-3}{6} = -\frac{1}{2} $

But here it's $ \frac{1}{27} $, so not.

So perhaps leave as $ -\frac{1}{2} \log_2 3 $, but not clean.

Wait — maybe it's $ \log_{64} \frac{1}{64} $? Then answer is -1.

But it's $ \frac{1}{27} $.

So likely, this is not simplifiable.

But let's move on.

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18) $ \log_{\frac{1}{16}} 32 = ? $

Base: $ \frac{1}{16} = 2^{-4} $, Argument: $ 32 = 2^5 $

Let $ x = \log_{\frac{1}{16}} 32 $

Then:
$$
\left(\frac{1}{16}\right)^x = 32 \Rightarrow (2^{-4})^x = 2^{-4x} = 2^5 \Rightarrow -4x = 5 \Rightarrow x = -\frac{5}{4}
$$
✔ Answer: $ -\frac{5}{4} $

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19) $ \log_4 0 = ? $

Logarithm of 0 is undefined (no power of 4 gives 0)

✔ Answer: undefined

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20) $ \log_9 1 = ? $

Any log of 1 is 0, because $ b^0 = 1 $

✔ Answer: 0

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21) $ \log_2 \frac{1}{8} = ? $

$ \frac{1}{8} = 2^{-3} \Rightarrow \log_2 (2^{-3}) = -3 $

✔ Answer: -3

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22) $ \log_{27} \frac{1}{3} = ? $

$ 27 = 3^3 $, $ \frac{1}{3} = 3^{-1} $

Let $ x = \log_{27} \frac{1}{3} $

Then:
$$
27^x = \frac{1}{3} \Rightarrow (3^3)^x = 3^{3x} = 3^{-1} \Rightarrow 3x = -1 \Rightarrow x = -\frac{1}{3}
$$
✔ Answer: $ -\frac{1}{3} $

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23) $ \log_4 3 = ? $

This is not a nice number. Can't simplify without calculator.

But we can write:
$$
\log_4 3 = \frac{\log 3}{\log 4} = \frac{\log 3}{2\log 2}
$$

But no exact value.

So likely, this is just left as $ \log_4 3 $, or perhaps they want us to recognize it's not simplifiable.

But let's see if it's part of a pattern.

Wait — maybe it's $ \log_4 16 $? Then 2.

But it's $ \log_4 3 $, so probably not.

So answer is just $ \log_4 3 $, but not numerical.

But maybe it's expected to be left as is.

Wait — no, it asks to evaluate.

But $ \log_4 3 $ is irrational, so can't be simplified.

So likely, not evaluable exactly without calculator.

But perhaps it's $ \log_4 4 $? Then 1.

But it's $ \log_4 3 $, so leave as is.

But maybe we can write in terms of other logs.

But the instruction says "evaluate", so perhaps it's a trick.

Wait — perhaps it's $ \log_4 4 $, but no.

So I think this one is not simplifiable.

But let's continue.

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24) $ \log_6 6^x = ? $

By log rule: $ \log_b (b^x) = x $

So $ \log_6 (6^x) = x $

✔ Answer: $ x $

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25) $ \log_{36} \frac{1}{6} = ? $

$ 36 = 6^2 $, $ \frac{1}{6} = 6^{-1} $

Let $ x = \log_{36} \frac{1}{6} $

Then:
$$
36^x = \frac{1}{6} \Rightarrow (6^2)^x = 6^{2x} = 6^{-1} \Rightarrow 2x = -1 \Rightarrow x = -\frac{1}{2}
$$
✔ Answer: $ -\frac{1}{2} $

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26) $ \log_{128} 2 = ? $

$ 128 = 2^7 $, so:
$$
\log_{2^7} 2 = \frac{\log 2}{\log (2^7)} = \frac{\log 2}{7 \log 2} = \frac{1}{7}
$$
✔ Answer: $ \frac{1}{7} $

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27) $ \log_{\frac{1}{4}} 16 = ? $

$ \frac{1}{4} = 4^{-1} = (2^2)^{-1} = 2^{-2} $, $ 16 = 2^4 $

Let $ x = \log_{\frac{1}{4}} 16 $

Then:
$$
\left(\frac{1}{4}\right)^x = 16 \Rightarrow (4^{-1})^x = 4^{-x} = 16
\Rightarrow (2^2)^{-x} = 2^{-2x} = 2^4 \Rightarrow -2x = 4 \Rightarrow x = -2
$$
✔ Answer: -2

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28) $ \log_8 x^2 = ? $

This is not a number — depends on $ x $. But it's written as an expression.

But the problem says "evaluate", so likely it's asking for a simplified expression.

Using log rules:
$$
\log_8 x^2 = 2 \log_8 x
$$

Or, if $ x > 0 $, it's $ 2 \log_8 x $

But unless $ x $ is given, can't evaluate numerically.

But perhaps it's $ \log_8 8^2 $? Then 2.

But it's $ x^2 $, so likely the answer is $ 2 \log_8 x $

But the instruction says "evaluate", so maybe it's a typo.

Wait — perhaps it's $ \log_8 64 $? $ 64 = 8^2 $, so $ \log_8 64 = 2 $

But it's $ x^2 $, so probably $ 2 \log_8 x $

So Answer: $ 2 \log_8 x $

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29) $ \ln e^{12} = ? $

$ \ln(e^{12}) = 12 \cdot \ln e = 12 \cdot 1 = 12 $

✔ Answer: 12

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30) $ 3^{\log_3 5} = ? $

Use identity: $ b^{\log_b a} = a $

So $ 3^{\log_3 5} = 5 $

✔ Answer: 5

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31) $ \ln 1 = ? $

$ \ln 1 = 0 $, because $ e^0 = 1 $

✔ Answer: 0

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32) $ e^{\ln 4} = ? $

$ e^{\ln 4} = 4 $, because $ e^{\ln x} = x $

✔ Answer: 4

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Now go back to the ones we skipped.

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13) $ \log_{\frac{1}{3}} \frac{9}{4} = ? $

Let’s try to write in exponential form.

Let $ x = \log_{\frac{1}{3}} \frac{9}{4} $

Then:
$$
\left(\frac{1}{3}\right)^x = \frac{9}{4}
\Rightarrow 3^{-x} = \frac{9}{4} = \frac{3^2}{2^2}
\Rightarrow 3^{-x} = 3^2 \cdot 2^{-2}
$$

Take log base 3:
$$
\log_3 (3^{-x}) = \log_3 (3^2 \cdot 2^{-2}) = \log_3 3^2 + \log_3 2^{-2} = 2 - 2\log_3 2
\Rightarrow -x = 2 - 2\log_3 2
\Rightarrow x = -2 + 2\log_3 2
$$

But this is not a nice number.

Alternatively, write as:
$$
x = \log_{\frac{1}{3}} \frac{9}{4} = \frac{\log \frac{9}{4}}{\log \frac{1}{3}} = \frac{\log 9 - \log 4}{\log 1 - \log 3} = \frac{2\log 3 - 2\log 2}{-\log 3} = \frac{2(\log 3 - \log 2)}{-\log 3} = -2\left(1 - \frac{\log 2}{\log 3}\right)
$$

Still messy.

So likely, this one is not meant to be evaluated numerically, or there's a typo.

But perhaps the argument is $ \frac{1}{9} $? Then:
$$
\left(\frac{1}{3}\right)^x = \frac{1}{9} = \left(\frac{1}{3}\right)^2 \Rightarrow x = 2
$$

That makes sense.

Similarly, for 17) $ \log_{64} \frac{1}{27} $ — if it were $ \frac{1}{64} $, then $ x = -1 $

But as is, not simplifiable.

So perhaps assume typos.

But let's assume the problems are correct.

For 13), since it's not simplifiable, maybe it's $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) = 2 $, but written as $ \frac{9}{4} $ by mistake.

Alternatively, perhaps $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) $, but it's $ \frac{9}{4} $.

Given the context, maybe it's $ \log_{\frac{1}{3}} \left(\frac{1}{9}\right) $, so answer is 2.

But we'll go with what's written.

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17) $ \log_{64} \frac{1}{27} $

As before, $ 64 = 2^6 $, $ 27 = 3^3 $, so no common base.

So $ \log_{64} \frac{1}{27} = \frac{\log \frac{1}{27}}{\log 64} = \frac{-\log 27}{\log 64} = \frac{-3\log 3}{6\log 2} = -\frac{1}{2} \cdot \frac{\log 3}{\log 2} = -\frac{1}{2} \log_2 3 $

Not a nice number.

So likely, either typo or not evaluable.

But perhaps it's $ \log_{64} \frac{1}{64} = -1 $

But it's $ \frac{1}{27} $.

So we'll leave it.

But let's assume the problems are as written.

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Final Answers:

| Problem | Answer |
|--------|--------|
| 9) $ \log_3 81 $ | 4 |
| 10) $ \log_4 \frac{1}{2} $ | $ -\frac{1}{2} $ |
| 11) $ \log_{12} 144 $ | 2 |
| 12) $ \log_6 \frac{1}{36} $ | -2 |
| 13) $ \log_{\frac{1}{3}} \frac{9}{4} $ | $ -2 + 2\log_3 2 $ (or not simplifiable) |
| 14) $ \log_{0.25} 4 $ | -1 |
| 15) $ \log_3 (-3) $ | undefined |
| 16) $ \log_4 4 $ | 1 |
| 17) $ \log_{64} \frac{1}{27} $ | $ -\frac{1}{2} \log_2 3 $ (or not simplifiable) |
| 18) $ \log_{\frac{1}{16}} 32 $ | $ -\frac{5}{4} $ |
| 19) $ \log_4 0 $ | undefined |
| 20) $ \log_9 1 $ | 0 |
| 21) $ \log_2 \frac{1}{8} $ | -3 |
| 22) $ \log_{27} \frac{1}{3} $ | $ -\frac{1}{3} $ |
| 23) $ \log_4 3 $ | $ \log_4 3 $ (cannot simplify) |
| 24) $ \log_6 6^x $ | $ x $ |
| 25) $ \log_{36} \frac{1}{6} $ | $ -\frac{1}{2} $ |
| 26) $ \log_{128} 2 $ | $ \frac{1}{7} $ |
| 27) $ \log_{\frac{1}{4}} 16 $ | -2 |
| 28) $ \log_8 x^2 $ | $ 2 \log_8 x $ |
| 29) $ \ln e^{12} $ | 12 |
| 30) $ 3^{\log_3 5} $ | 5 |
| 31) $ \ln 1 $ | 0 |
| 32) $ e^{\ln 4} $ | 4 |

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Notes:

- Problems 13, 17, 23 involve non-integer or irrational answers that require logs.
- Problem 28 is an expression, not a number.
- Problems 15 and 19 are undefined.
- Others are standard.

If you believe there are typos (e.g., 13 should be $ \frac{1}{9} $, 17 should be $ \frac{1}{64} $), then answers would be simpler.

But based on what's written, above is the best evaluation.

Let me know if you'd like clarification on any!