Logarithms - The Easy Way! - Free Printable

You're given six logarithmic expressions to evaluate. Let’s solve each one step by step and explain the reasoning.

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1. log₂(16)

This asks: *“To what power must 2 be raised to get 16?”*

We know:
> 2⁴ = 16

✔ So, log₂(16) = 4

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2. log₃(27)

This asks: *“To what power must 3 be raised to get 27?”*

We know:
> 3³ = 27

✔ So, log₃(27) = 3

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3. log(1000)

When no base is written, it means base 10 (common logarithm).

So this is: log₁₀(1000)

Ask: *“To what power must 10 be raised to get 1000?”*

> 10³ = 1000

✔ So, log(1000) = 3

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4. log₃(1/9)

This asks: *“To what power must 3 be raised to get 1/9?”*

Note that:
> 1/9 = 3⁻²

Because 3² = 9 → 3⁻² = 1/9

✔ So, log₃(1/9) = -2

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5. 5^(log₅(14))

This is a classic inverse function property.

Recall the identity:
> a^(logₐ(x)) = x, for a > 0, a ≠ 1, x > 0

Here, base is 5, and we’re raising 5 to the power of log₅(14). So:

✔ 5^(log₅(14)) = 14

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6. log₉(1/3)

This asks: *“To what power must 9 be raised to get 1/3?”*

Let’s write everything in terms of base 3:

- 9 = 3²
- 1/3 = 3⁻¹

So let x = log₉(1/3)

Then by definition:
> 9^x = 1/3
> (3²)^x = 3⁻¹
> 3^(2x) = 3⁻¹

Since bases are equal, set exponents equal:
> 2x = -1
> x = -1/2

✔ So, log₉(1/3) = -1/2

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## ✔ Final Answers:

1. log₂(16) = 4
2. log₃(27) = 3
3. log(1000) = 3
4. log₃(1/9) = -2
5. 5^(log₅(14)) = 14
6. log₉(1/3) = -1/2

These solutions use basic logarithmic definitions, exponent rules, and inverse properties. Great practice!