What is a logarithm? The exponent you’re looking for

By Marius, private math tutor in Montreal · 2026-10-05

A logarithm is an exponent: $\log_{c} x$ is the power you have to put on the base $c$ to get $x$ , so $\log_{2} 16 = 4$ because $2^{4} = 16$ . Everything else about logarithms, the three laws included, follows from that one sentence and the laws of exponents.

The question a logarithm answers

Some exponential equations can be solved by eye. $2^{x} = 16$ has the solution $x = 4$ , because you know your powers of 2. Now try $2^{x} = 10$ .

$2^{3} = 8$ is too small and $2^{4} = 16$ is too big, so the solution lies between 3 and 4.
No fraction is exactly equal to it. So instead of hunting for one, mathematics gives the number a name, the same way it named $\sqrt{2}$ .
The name is $\log_{2} 10$ , read “log base 2 of 10”: the exponent that turns 2 into 10.

In general, for a base $c > 0$ with $c \neq 1$ , and a number $x > 0$ :

\log_{c} x = y ⟺ c^{y} = x

$c$ is the base, and it is the same base in both forms.
$x$ is the argument, the number you are trying to reach.
$y$ , the value of the logarithm, is the exponent.
$\log x$ with no base written means base 10, the common logarithm. It is the log key on a calculator.

Switching between the two forms

The base stays the base, and the logarithm is always the exponent. Each line below says the same thing twice:

$\log_{3} 9 = 2$ and $3^{2} = 9$
$\log_{10} 1000 = 3$ and $10^{3} = 1000$
$\log_{2} (\frac{1}{8}) = - 3$ and $2^{- 3} = \frac{1}{8}$
$\log_{4} 2 = \frac{1}{2}$ and $4^{1 / 2} = 2$
$\log_{6} 1 = 0$ and $6^{0} = 1$

Three things to notice in that list:

A logarithm can be negative. With a base greater than 1, that happens when the argument is between 0 and 1, which takes a negative exponent to reach.
A logarithm can be a fraction: $4^{1 / 2}$ is the square root of 4.
The argument can never be zero or negative, because a positive base raised to any power gives a positive result. A logarithm only accepts arguments in $] 0, + \infty [$ .

If you are unsure which number goes where, say it out loud: “log base 3 of 9 is 2” becomes “3 to the power 2 is 9.” The small number written in the subscript is the one being raised to a power.

Evaluating a logarithm without a calculator

Rewrite the logarithm as an exponential equation, write both sides as powers of the same number, and compare the exponents.

Worked example 1: evaluate $\log_{8} 32$ .

Call the answer $y$ . In exponential form, $8^{y} = 32$ .
32 is not a whole-number power of 8, but both are powers of 2: $8 = 2^{3}$ and $32 = 2^{5}$ . So ${(2^{3})}^{y} = 2^{5}$ , that is, $2^{3 y} = 2^{5}$ .
Two equal powers of the same base have equal exponents: $3 y = 5$ , so $y = \frac{5}{3}$ .
Check: $8^{5 / 3} = {(\sqrt[3]{8})}^{5} = 2^{5} = 32$ .

Worked example 2: solve $4^{x} = 11$ exactly.

4 and 11 are not powers of a common whole number, so no common-base trick will work. Going through the definition the other way does: the exponent that turns 4 into 11 is, by definition, $\log_{4} 11$ . So $x = \log_{4} 11$ . That is the exact answer, and since $4^{1} = 4$ and $4^{2} = 16$ , it lies between 1 and 2.

Where the three laws come from

The laws of logarithms are not new rules to memorise. Each one is a law of exponents, translated through the definition. Take two positive numbers $M$ and $N$ , and name their logarithms:

m = \log_{c} M n = \log_{c} N

By the definition, that says exactly the same thing as

M = c^{m} N = c^{n}

The product law

Multiply. Multiplying powers of the same base adds the exponents:

M N = c^{m} \cdot c^{n} = c^{m + n}

So the exponent that turns $c$ into $M N$ is $m + n$ . Written as a logarithm:

\log_{c} (M N) = \log_{c} M + \log_{c} N

The quotient law

Divide. Dividing powers of the same base subtracts the exponents:

\frac{M}{N} = \frac{c^{m}}{c^{n}} = c^{m - n} so \log_{c} (\frac{M}{N}) = \log_{c} M - \log_{c} N

The power law

Raise $M$ to any power $p$ . A power of a power multiplies the exponents:

M^{p} = {(c^{m})}^{p} = c^{m p} so \log_{c} (M^{p}) = p \log_{c} M

Two special values come from the same place: $c^{0} = 1$ gives $\log_{c} 1 = 0$ , and $c^{1} = c$ gives $\log_{c} c = 1$ . As long as you remember the exponent laws, you can rebuild every one of these in a few lines whenever you doubt it.

Worked example 3: evaluate $\log_{6} 4 + \log_{6} 9$ and $\log_{2} \sqrt[3]{16}$ .

By the product law, $\log_{6} 4 + \log_{6} 9 = \log_{6} 36 = 2$ , since $6^{2} = 36$ . Neither logarithm is a tidy number on its own; their sum is.
Write the root as a power, $\sqrt[3]{16} = 16^{1 / 3}$ . By the power law, $\log_{2} (16^{1 / 3}) = \frac{1}{3} \log_{2} 16 = \frac{1}{3} (4) = \frac{4}{3}$ .

The laws cover a product, a quotient and a power, and nothing else. They exist because the exponent laws say what happens when powers of the same base are multiplied, divided or raised to a power. The exponent laws say nothing about adding two powers, so there is no law for the logarithm of a sum: $\log_{c} (M + N)$ does not split into $\log_{c} M$ and $\log_{c} N$ . The trap is to split it anyway, because it looks like the product law. When the argument is a sum, work out the sum first or leave it as it is.

What your calculator does with other bases

The log key on a calculator works in base 10, and the power law converts any other base to it. If $y = \log_{c} x$ , then $c^{y} = x$ . Take the common logarithm of both sides and bring the exponent down:

\log (c^{y}) = \log x \Rightarrow y \log c = \log x \Rightarrow \log_{c} x = \frac{\log x}{\log c}

For the opening question, $\log_{2} 10 = \frac{\log 10}{\log 2} = \frac{1}{\log 2} \approx 3.32$ : between 3 and 4, as predicted.

Check your understanding

No calculator.

Write $\log_{7} 343 = 3$ in exponential form, and $3^{- 3} = \frac{1}{27}$ in logarithmic form.
Evaluate $\log_{16} 8$ .
True or false: $\log_{5} 0 = 0$ . Explain.
Between which two consecutive whole numbers does $\log_{5} 400$ lie? Explain how you know.
Solve $\log_{2} (3 x - 1) = 5$ .
Write $\log_{7} 14 + \log_{7} 21 - \log_{7} 6$ as a single logarithm, then evaluate it.
Find the error. A student evaluates $\log_{3} (9 \cdot 27)$ by multiplying $\log_{3} 9$ by $\log_{3} 27$ , and gets 6. What went wrong, and what is the correct value?
Use the definition and a law of exponents to show that $\log_{c} (\frac{1}{x}) = - \log_{c} x$ for every base $c > 0$ , $c \neq 1$ , and every $x > 0$ .

Show the answers

$7^{3} = 343$ and $\log_{3} (\frac{1}{27}) = - 3$ . The base stays the base in both forms, and the logarithm is the exponent.
$\frac{3}{4}$ . Write $16^{y} = 8$ in base 2: $2^{4 y} = 2^{3}$ , so $4 y = 3$ . Check: $16^{3 / 4} = {(\sqrt[4]{16})}^{3} = 2^{3} = 8$ .
False. $\log_{5} 0$ would be an exponent $y$ with $5^{y} = 0$ , and no power of 5 is 0, so $\log_{5} 0$ does not exist. The logarithm that equals 0 is $\log_{5} 1$ , because $5^{0} = 1$ .
Between 3 and 4. $5^{3} = 125$ and $5^{4} = 625$ , and 400 lies between them. Powers of 5 increase with the exponent, so the exponent that gives 400 is between 3 and 4.
$x = 11$ . In exponential form, $3 x - 1 = 2^{5} = 32$ , so $3 x = 33$ . The argument $3 (11) - 1 = 32$ is positive, so the solution stands.
$\log_{7} 49 = 2$ . The product law, then the quotient law, give $\log_{7} (\frac{14 \cdot 21}{6}) = \log_{7} (\frac{294}{6}) = \log_{7} 49$ , and $7^{2} = 49$ .
The product law turns the logarithm of a product into a sum of logarithms, not a product of them. Correctly, $\log_{3} 9 + \log_{3} 27 = 2 + 3 = 5$ . Check: $9 \cdot 27 = 243 = 3^{5}$ .
Let $y = \log_{c} x$ , so $c^{y} = x$ . By the law of negative exponents, $\frac{1}{x} = \frac{1}{c^{y}} = c^{- y}$ . So the exponent that turns $c$ into $\frac{1}{x}$ is $- y$ , which means $\log_{c} (\frac{1}{x}) = - y = - \log_{c} x$ .

To practise, the free Secondary 4 Arithmetic worksheet has a sheet on evaluating logarithms and one on their laws, and the Secondary 4 Logarithmic Functions worksheet takes the same definition into graphs, domains and inverses.

Want one-on-one help with this? I offer Secondary 4 and 5 math tutoring in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.