EL2, EL3 - Logarithms and their laws

A logarithm answers a missing-power question. EL2 turns powers into logarithms and back; EL3 explains why products become sums and powers become multipliers. These laws will let you solve equations and reveal straight lines inside curved data.

A logarithm asks for an exponent

Why does adding $\log_{10}2$ to $\log_{10}5$ give exactly $1$ , although neither number is a whole number? The answer will come from multiplying $2$ by $5$ , but first we need to understand what a logarithm means.

You already know

For a positive base, $a^0=1$ , $a^{-r}=1/a^r$ , and $a^{u+v}=a^ua^v$ . If $a\ne1$ , the exponential function rises or falls throughout, so it takes each positive output once.

The question “What power of $2$ gives $8$ ?” has answer $3$ , because $2^3=8$ . We write that answer as $\log_2 8=3$ . The small $2$ is the base; $8$ is the input, also called the argument. The logarithm is the exponent, not the result of dividing $8$ by $2$ .

Logarithm

$\log_a x$ is the power to which $a$ must be raised to obtain $x$ . Thus $\log_a x=u$ means $a^u=x$ . Its real-number conditions are $a>0$ , $a\ne1$ and $x>0$ . For example, $\log_3 9=2$ ; $\log_3(-9)$ has no real value.

A logarithm is the inverse of an exponential: it undoes raising a fixed base to a power. Base $1$ cannot work because every power of $1$ is $1$ , so the output does not identify one exponent. Zero and negative arguments cannot work because powers of a positive base are positive.

Exponential and logarithmic forms

\log_a x=u\quad\Longleftrightarrow\quad a^u=x

For $a>0$ , $a\ne1$ and $x>0$ , this gives $\log_a1=0$ , $\log_a a=1$ , $\log_a(a^u)=u$ and $a^{\log_a x}=x$ .
Not on the formulae sheet: learn it.

Base $10$ is especially useful. Its logarithms count powers of $10$ : $\log_{10}1000=3$ , while $\log_{10}0.001=-3$ . A negative logarithm is perfectly possible; a logarithm of a negative number is a different matter.

Key idea

Ask “What exponent?” A logarithm can be negative, zero or positive, but its argument must be positive.

Separate the sign of the answer from the sign of the argument.

Evaluating logarithms from powers

Before reaching for a calculator, see whether the argument is a familiar power of the base.

Convert the logarithm into a missing-power equation. Fractions can give negative exponents; roots can give fractional exponents. Neither changes the meaning of the logarithm.

Worked example 1: a fractional exponent

Question. Find the exact value of $\log_9 27$ .

Plan. Write both numbers as powers of $3$ , then equate exponents.

\begin{aligned} \log_9 27=u&\Longrightarrow9^u=27 && \text{definition of logarithm}\\ (3^2)^u&=3^3 && \text{common base}\\ 2u&=3 && \text{equal powers of }3\\ u&=\frac32 && \text{divide by }2 \end{aligned}

Why base $3$ ? Both $9$ and $27$ are exact powers of $3$ , so the unknown moves into an ordinary linear equation.

Answer. $\log_9 27=\frac32$ .

Check. $9^{3/2}=(\sqrt9)^3=3^3=27$ .

Change only the argument to its reciprocal. The required exponent changes sign.

Worked example 2: a negative fractional exponent

Question. Find the exact value of $\log_9(\frac1{27})$ .

Plan. Use the same common base, now with a negative power for the argument.

\begin{aligned} 9^u&=\frac1{27} && \text{logarithmic definition}\\ 3^{2u}&=3^{-3} && \text{write reciprocal as a power}\\ 2u&=-3 && \text{equate exponents}\\ u&=-\frac32 && \text{divide by }2 \end{aligned}

Why is a negative answer allowed? The argument $1/27$ is positive; the negative exponent describes a reciprocal.

Answer. $\log_9(\frac1{27})=-\frac32$ .

Check. $9^{-3/2}=1/27$ , exactly.

The same reasoning works for bases between $0$ and $1$ , though the direction reverses: for example, $(\frac12)^{-3}=8$ , so $\log_{1/2}8=-3$ .

Watch out

Wrong: $\log_9 27=27/9=3$ . A logarithm is not division. Right: solve $9^u=27$ , giving $u=3/2$ .

Try another reciprocal while keeping the common-base method.

Products and quotients inside logarithms

Multiplying numbers adds their exponents, which is why multiplying arguments adds their logarithms.

Let $u=\log_a x$ and $v=\log_a y$ , so $x=a^u$ and $y=a^v$ . Then $xy=a^{u+v}$ , hence $\log_a(xy)=u+v$ . Similarly, $x/y=a^{u-v}$ , so its logarithm is $u-v$ . The laws come from indices rather than a new calculator trick.

Product and quotient laws

\log_a x+\log_a y=\log_a(xy)

\log_a x-\log_a y=\log_a\left(\frac{x}{y}\right)

All logarithms have the same base $a>0$ , $a\ne1$ , with $x>0$ and $y>0$ .
Not on the formulae sheet: learn it.

The laws can be read in either direction. Combining gives one logarithm; expanding gives a sum or difference.

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EL2, EL3 - Logarithms and their laws

A logarithm asks for an exponent

Evaluating logarithms from powers

Products and quotients inside logarithms

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