Introduction to Logarithms

How many $2$s do you multiply together to get $32$? The answer, $5$, has a name: it’s the logarithm of $32$ with base $2$. Logarithms answer the question “what exponent do I need?”, which is exactly what you need to solve equations like $3^x = 10$, where the unknown is stuck in the exponent. In Grade 11 you could only guess and check; logarithms give you a real method.

Key ideas

A logarithm is an exponent

For a base $b \gt 0$ with $b \ne 1$, and a number $x \gt 0$:

$$\log_b x = y \quad \Longleftrightarrow \quad b^y = x$$

Read $\log_b x$ as “log base $b$ of $x$”. It means the exponent you put on $b$ to get $x$. For example, $\log_2 8 = 3$ because $2^3 = 8$.

Exponential form and logarithmic form

The same fact can be written two ways. The base stays the base, and the logarithm is the exponent.

Exponential form	Logarithmic form
$5^3 = 125$	$\log_5 125 = 3$
$10^{-2} = 0.01$	$\log_{10} 0.01 = -2$
$9^{\frac{1}{2}} = 3$	$\log_9 3 = \tfrac{1}{2}$
$b^y = x$	$\log_b x = y$

Logarithms undo exponentiation

Taking a logarithm is the inverse operation of raising to a power, the same way a square root undoes squaring. If you raise $b$ to a power and then take $\log_b$, you get the power back:

$$\log_b\left(b^x\right) = x \qquad \text{and} \qquad b^{\log_b x} = x \ \ (x \gt 0)$$

For example, $\log_3\left(3^7\right) = 7$ and $10^{\log_{10} 42} = 42$. The logarithmic functions page shows what this looks like on a graph.

Why you can’t take the log of zero or a negative number

A positive base raised to any power gives a positive answer: $2^3 = 8$, $2^0 = 1$, $2^{-3} = \tfrac{1}{8}$. No exponent makes $2^y$ equal to $0$ or a negative number. So:

• $\log_2 0$ is undefined, because $2^y = 0$ has no solution.
• $\log_{10}(-3)$ is undefined, because no power of $10$ is negative.

The answer to a logarithm can be negative ($\log_{10} 0.01 = -2$), but the number you take the log of must be positive.

The base has rules too. It must be positive (powers of a negative number jump between positive and negative), and it can’t be $1$ (every power of $1$ is $1$, so $\log_1 7$ would have no answer).

The common logarithm

A logarithm with base $10$ is called a common logarithm, and it’s usually written without the base:

$$\log x \quad \text{means} \quad \log_{10} x$$

So $\log 1000 = 3$ and $\log 0.1 = -1$. This is the LOG key on your calculator.

Estimating logarithms

Most logarithms aren’t whole numbers. To estimate one, find the powers of the base on either side.

For $\log_2 50$: since $2^5 = 32$ and $2^6 = 64$, and $32 \lt 50 \lt 64$, the value is between $5$ and $6$. Then narrow it down by systematic trial on a calculator (Example 3).

For base $10$, the LOG key gives the value directly: $\log 50 \approx 1.699$. For other bases, there’s a shortcut called the change of base formula, which you’ll meet with the laws of logarithms.

Worked examples

Example 1: Evaluating simple logarithms

Evaluate each logarithm.

• (a) $\log_2 64$
• (b) $\log_3 \dfrac{1}{9}$
• (c) $\log 0.001$
• (d) $\log_{25} 5$
• (e) $\log_7 1$

Solution. For each one, ask “what power of the base gives this number?”

(a) $2^6 = 64$, so $\log_2 64 = 6$.

(b) $3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$, so $\log_3 \dfrac{1}{9} = -2$.

(c) $0.001 = 10^{-3}$, so $\log 0.001 = -3$.

(d) $25^{\frac{1}{2}} = \sqrt{25} = 5$, so $\log_{25} 5 = \dfrac{1}{2}$.

(e) $7^0 = 1$, so $\log_7 1 = 0$. (In fact, $\log_b 1 = 0$ for every base.)

Example 2: Switching forms

(a) Write in logarithmic form: $4^3 = 64$ and $8^{\frac{2}{3}} = 4$.

(b) Write in exponential form: $\log_6 216 = 3$ and $\log 1\,000\,000 = 6$.

(c) Solve $\log_2 x = 5$ and $\log_x 49 = 2$.

Solution.

(a) The exponent becomes the value of the log: $\log_4 64 = 3$ and $\log_8 4 = \dfrac{2}{3}$.

(b) The value of the log becomes the exponent: $6^3 = 216$ and $10^6 = 1\,000\,000$.

(c) Rewrite each in exponential form.

$$\log_2 x = 5 \quad\Rightarrow\quad x = 2^5 = 32$$ $$\log_x 49 = 2 \quad\Rightarrow\quad x^2 = 49 \quad\Rightarrow\quad x = 7$$

We reject $x = -7$ because a base must be positive. Check: $7^2 = 49$. ✓

Example 3: Estimating by systematic trial

Estimate $\log_2 50$ to two decimal places.

Solution. We want the exponent $y$ with $2^y = 50$. From the powers $2^5 = 32$ and $2^6 = 64$, $y$ is between $5$ and $6$. Since $50$ is a bit closer to $64$ (in ratio), try a value past the middle.

Try $y$	$2^y$	Too big or too small?
$5.6$	$48.50$	too small
$5.7$	$51.98$	too big
$5.64$	$49.87$	too small
$5.65$	$50.21$	too big

So $y$ is between $5.64$ and $5.65$, and $2^{5.64}$ is closer to $50$. To two decimal places, $\log_2 50 \approx 5.64$.

Example 4: Solving by rewriting in log form

Solve $3^x = 10$. Give an exact answer, then an approximation to two decimal places.

Solution. Rewrite in logarithmic form. The base is $3$ and the exponent is $x$:

$$3^x = 10 \quad\Longleftrightarrow\quad x = \log_3 10$$

That’s the exact answer. To estimate it: $3^2 = 9$ and $3^3 = 27$, so $x$ is a little more than $2$.

Try $x$	$3^x$
$2.1$	$10.05$ (too big)
$2.09$	$9.94$ (too small)
$2.095$	$9.99$ (too small)

So $x$ is between $2.095$ and $2.1$, which rounds to $x \approx 2.10$. Check: $3^{2.10} \approx 10.05 \approx 10$. ✓

Common mistakes

Mixing up which number is the exponent. $\log_2 8$ is not $2^8$ or $8^2$. It asks ”$2$ to what power is $8$?”, so the answer is $3$. Say the question out loud until it’s automatic.

Taking the log of zero or a negative number. $\log(-100)$ and $\log_5 0$ don’t exist, because a positive base raised to any power is positive. A calculator will show an error.

Thinking a log can’t be negative. It can. $\log 0.01 = -2$ because $10^{-2} = 0.01$. Logs of numbers between $0$ and $1$ are negative (for bases bigger than $1$).

Forgetting what “log” with no base means. $\log x$ is base $10$. So $\log 100 = 2$, not $100$ or $50$.

Treating the log like multiplication. $\log_2 50$ is not $\dfrac{50}{2}$ or $2 \times 50$. It’s the exponent that turns $2$ into $50$, about $5.64$.

Getting fractional logs backwards. $\log_9 3 = \tfrac{1}{2}$ (because $9^{\frac{1}{2}} = 3$), but $\log_3 9 = 2$. When the number is smaller than the base (and bigger than $1$), the log is between $0$ and $1$.

Practice

1. (Warm-up) Evaluate.

• (a) $\log_5 25$
• (b) $\log_2 \dfrac{1}{8}$
• (c) $\log 10\,000$

Solution

(a) $5^2 = 25$, so $\log_5 25 = 2$.

(b) $2^{-3} = \dfrac{1}{8}$, so $\log_2 \dfrac{1}{8} = -3$.

(c) $10^4 = 10\,000$, so $\log 10\,000 = 4$.

2. (Warm-up) Write in logarithmic form.

• (a) $3^4 = 81$
• (b) $10^{-3} = 0.001$
• (c) $16^{\frac{1}{4}} = 2$

Solution

(a) $\log_3 81 = 4$

(b) $\log 0.001 = -3$

(c) $\log_{16} 2 = \dfrac{1}{4}$

3. (Warm-up) Write in exponential form.

• (a) $\log_7 343 = 3$
• (b) $\log_4 \dfrac{1}{16} = -2$
• (c) $\log 1 = 0$

Solution

(a) $7^3 = 343$

(b) $4^{-2} = \dfrac{1}{16}$

(c) $10^0 = 1$

4. (Core) Evaluate.

• (a) $\log_8 2$
• (b) $\log_{27} 9$
• (c) $\log_{\frac{1}{2}} 8$
• (d) $\log_5 \sqrt{5}$

Solution

(a) $8^{\frac{1}{3}} = \sqrt[3]{8} = 2$, so $\log_8 2 = \dfrac{1}{3}$.

(b) $27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$, so $\log_{27} 9 = \dfrac{2}{3}$.

(c) $\left(\dfrac{1}{2}\right)^{-3} = 2^3 = 8$, so $\log_{\frac{1}{2}} 8 = -3$.

(d) $\sqrt{5} = 5^{\frac{1}{2}}$, so $\log_5 \sqrt{5} = \dfrac{1}{2}$.

5. (Core) Solve each equation by rewriting it in exponential form.

• (a) $\log_3 x = 4$
• (b) $\log_x 64 = 3$
• (c) $\log_2 (x - 1) = 5$
• (d) $\log x = -2$

Solution

(a) $x = 3^4 = 81$

(b) $x^3 = 64$, so $x = \sqrt[3]{64} = 4$.

(c) $x - 1 = 2^5 = 32$, so $x = 33$. Check: $\log_2 32 = 5$. ✓

(d) $x = 10^{-2} = 0.01$

6. (Core) Between which two consecutive integers is each logarithm? Explain.

• (a) $\log_4 50$
• (b) $\log 0.05$
• (c) $\log_6 200$

Solution

(a) $4^2 = 16$ and $4^3 = 64$. Since $16 \lt 50 \lt 64$, $\log_4 50$ is between $2$ and $3$.

(b) $10^{-2} = 0.01$ and $10^{-1} = 0.1$. Since $0.01 \lt 0.05 \lt 0.1$, $\log 0.05$ is between $-2$ and $-1$.

(c) $6^2 = 36$ and $6^3 = 216$. Since $36 \lt 200 \lt 216$, $\log_6 200$ is between $2$ and $3$ (and close to $3$).

7. (Core) Solve $2^x = 20$ by rewriting it in logarithmic form. Then estimate $x$ to two decimal places by systematic trial.

Solution

$x = \log_2 20$. Since $2^4 = 16$ and $2^5 = 32$, $x$ is between $4$ and $5$.

Try $x$	$2^x$
$4.3$	$19.70$ (too small)
$4.4$	$21.11$ (too big)
$4.32$	$19.97$ (too small)
$4.33$	$20.11$ (too big)

$2^{4.32}$ is much closer to $20$, so $x \approx 4.32$.

8. (Challenge) Explain why each of these is undefined.

• (a) $\log_5 0$
• (b) $\log_1 7$
• (c) $\log_{10}(-10)$

Solution

(a) We’d need $5^y = 0$. But every power of $5$ is positive (for example, $5^{-10}$ is tiny but still positive), so there’s no such $y$.

(b) We’d need $1^y = 7$. But $1$ to any power is $1$, so there’s no such $y$. That’s why the base can’t be $1$.

(c) We’d need $10^y = -10$. Every power of $10$ is positive, so there’s no such $y$.

9. (Challenge) Evaluate.

• (a) $\log_2\left(\log_3 81\right)$
• (b) $5^{\log_5 12}$
• (c) $\log_3\left(9^4\right)$

Solution

(a) Work from the inside out. $\log_3 81 = 4$, because $3^4 = 81$. Then $\log_2 4 = 2$.

(b) $\log_5 12$ is the exponent that turns $5$ into $12$. Putting that exponent on $5$ gives $12$. So $5^{\log_5 12} = 12$.

(c) $9^4 = \left(3^2\right)^4 = 3^8$, so $\log_3\left(9^4\right) = \log_3\left(3^8\right) = 8$.