Logarithms and the logarithm laws

A logarithm is an exponent. For a valid base $a$,

$$\boxed{\log_a x=y\quad\Longleftrightarrow\quad a^y=x}, \qquad a>0,\quad a\ne1,\quad x>0.$$

For example, $\log_2 8=3$ because $2^3=8$. This definition is the key to the whole topic. It explains logarithm values, their laws, their graphs and their use in solving exponential equations.

Before you begin

You should be able to:

• use positive, negative, zero and fractional indices;
• understand inverse functions from composite and inverse functions;
• recognise the graphs and restrictions of exponential functions.

Reading a logarithm as a question

To evaluate $\log_a x$, ask:

To what power must I raise $a$ to obtain $x$?

Worked example 1: integer, negative and fractional answers

Evaluate $\log_3 81$, $\log_5(1/125)$ and $\log_{16}2$.

Translate each logarithm into an exponential statement:

$$3^4=81 \quad\Longrightarrow\quad \boxed{\log_3 81=4},$$ $$5^{-3}=\frac1{125} \quad\Longrightarrow\quad \boxed{\log_5\left(\frac1{125}\right)=-3},$$

and

$$16^{1/4}=2 \quad\Longrightarrow\quad \boxed{\log_{16}2=\frac14}.$$

A negative logarithm is perfectly valid. It means the argument is produced by a negative power of the base. The logarithm itself may be any real number, although its argument must be positive.

Two values follow immediately from the definition:

$$\boxed{\log_a1=0} \quad\text{because}\quad a^0=1,$$ $$\boxed{\log_a a=1} \quad\text{because}\quad a^1=a.$$

Self check 1

Evaluate exactly:

1. $\log_7 49$
2. $\log_2(1/16)$
3. $\log_{25}5$
4. $\log_6 1$

Answers

1. $2$, since $7^2=49$.
2. $-4$, since $2^{-4}=1/16$.
3. $1/2$, since $25^{1/2}=5$.
4. $0$, since $6^0=1$.

Bases and arguments

In real A level mathematics, $\log_a x$ requires

$$a>0,\qquad a\ne1,\qquad x>0.$$

These restrictions have different purposes.

• The base must be positive so that $a^y$ is defined as a real number for every real $y$.
• The base cannot be $1$ because $1^y=1$ for every $y$, so it cannot produce a one-to-one function with an inverse.
• The argument must be positive because $a^y>0$ for every real $y$. No real power of a positive base gives $0$ or a negative number.

Therefore $\log_a0$ and $\log_a(-3)$ have no real values. Do not interpret $\log_a0$ as $0$: in fact, $\log_a1=0$.

When a logarithm contains an algebraic expression, the whole argument must be positive. For example,

$$\ln(2x-5)\text{ is defined when }2x-5>0,$$

so its domain is $x>5/2$.

Common and natural logarithms

Two bases have special calculator notation:

$$\log x=\log_{10}x, \qquad \ln x=\log_e x,$$

where $e=2.7182818\ldots$.

The common logarithm uses base $10$. The natural logarithm uses base $e$ and is central in calculus and continuous growth. A logarithm without a displayed base usually means base $10$ in A level calculator work, but always follow the notation in the question.

The inverse relationships give

$$\ln(e^x)=x \quad\text{for every }x\in\mathbb R,$$

and

$$e^{\ln x}=x \quad\text{for }x>0.$$

The domain condition in the second identity matters: $\ln x$ must exist first.

Worked example 2: use inverse operations

Simplify

$$\ln(e^{-7}),\qquad e^{\ln 12},\qquad \ln\left(\frac1{e^3}\right).$$

Since logarithms and exponentials with the same base undo one another,

$$\ln(e^{-7})=-7,$$ $$e^{\ln12}=12,$$

and, after writing $1/e^3=e^{-3}$,

$$\ln\left(\frac1{e^3}\right)=\ln(e^{-3})=-3.$$

The logarithm laws

For $M>0$, $N>0$ and a valid base $a$,

$$\boxed{\log_a(MN)=\log_aM+\log_aN}$$ $$\boxed{\log_a\left(\frac MN\right)=\log_aM-\log_aN}$$ $$\boxed{\log_a(M^k)=k\log_aM}.$$

They are index laws viewed through an inverse function. To justify the product law, let

$$p=\log_aM,\qquad q=\log_aN.$$

Then $M=a^p$ and $N=a^q$, so

$$MN=a^pa^q=a^{p+q}.$$

Taking $\log_a$ gives

$$\log_a(MN)=p+q=\log_aM+\log_aN.$$

The quotient and power laws follow similarly from $a^p/a^q=a^{p-q}$ and $(a^p)^k=a^{pk}$.

There is no sum law

The laws split products, quotients and powers, not sums or differences:

$$\log_a(M+N)\ne\log_aM+\log_aN.$$

For example, $\log(10+10)=\log20$, whereas $\log10+\log10=2$.

Worked example 3: evaluate using the laws

Given that $\log_a2=p$ and $\log_a3=q$, express $\log_a72$ in terms of $p$ and $q$.

Factorise the argument:

$$72=2^3\times3^2.$$

Apply the product and power laws:

$$\begin{aligned} \log_a72 &=\log_a(2^3\times3^2)\\ &=\log_a(2^3)+\log_a(3^2)\\ &=3\log_a2+2\log_a3\\ &=\boxed{3p+2q}. \end{aligned}$$

Worked example 4: expand one logarithm

Expand

$$\ln\left(\frac{x^4\sqrt y}{z^3}\right), \qquad x>0,\ y>0,\ z>0.$$

Use the quotient law, then the product law, then the power law:

$$\begin{aligned} \ln\left(\frac{x^4\sqrt y}{z^3}\right) &=\ln(x^4\sqrt y)-\ln(z^3)\\ &=\ln(x^4)+\ln(y^{1/2})-\ln(z^3)\\ &=\boxed{4\ln x+\frac12\ln y-3\ln z}. \end{aligned}$$

The positivity conditions make every logarithm in the expanded form valid.

Worked example 5: combine into one logarithm

Write

$$2\ln x-\frac12\ln(x+1)+\ln3$$

as one logarithm, where $x>0$.

First move coefficients into powers:

$$\ln(x^2)-\ln((x+1)^{1/2})+\ln3.$$

Addition corresponds to multiplication and subtraction corresponds to division:

$$\boxed{\ln\left(\frac{3x^2}{\sqrt{x+1}}\right)}.$$

Do not move a coefficient inside by multiplication. In general,

$$k\ln x=\ln(x^k),$$

not $\ln(kx)$.

Self check 2

1. Expand $\log_a(25x^3/y)$ fully.
2. Write $3\ln p+2\ln q-\ln r$ as one logarithm.
3. Given $\log_a2=u$ and $\log_a5=v$, express $\log_a(20/\sqrt2)$ in terms of $u$ and $v$.

Assume all variables used as logarithm arguments are positive.

Answers

1. $2\log_a5+3\log_ax-\log_ay$.
2. $\displaystyle \ln\left(\frac{p^3q^2}{r}\right)$.
3. Since $20/\sqrt2=2^{3/2}\times5$, the answer is $\displaystyle \frac32u+v$.

Change of base

Calculators provide $\log$ and $\ln$, but not usually $\log_a$ for every base. The change of base formula is

$$\boxed{\log_a x=\frac{\log_bx}{\log_ba}},$$

where $b$ is any valid base. In particular,

$$\log_a x=\frac{\ln x}{\ln a}=\frac{\log x}{\log a}.$$

To derive it, let $y=\log_ax$, so $a^y=x$. Taking $\log_b$ of both sides gives

$$y\log_ba=\log_bx,$$

and rearranging produces the formula.

Worked example 6: calculate an unfamiliar base

Evaluate $\log_7 20$ to four significant figures.

$$\log_720=\frac{\ln20}{\ln7}=1.539\ldots$$

Therefore

$$\boxed{\log_720=1.539\text{ to four significant figures}}.$$

The answer is plausible because $7^1<20<7^2$, so $1<\log_720<2$.

A useful consequence is the reciprocal identity

$$\boxed{\log_a b=\frac1{\log_b a}}.$$

Logarithmic graphs

Since $y=\log_ax$ is the inverse of $y=a^x$, their graphs are reflections of one another in the line $y=x$. Therefore every graph $y=\log_ax$ has:

• domain $x>0$;
• range $\mathbb R$;
• $x$-intercept $(1,0)$ because $\log_a1=0$;
• vertical asymptote $x=0$;
• no $y$-intercept.

If $a>1$, the graph is increasing. If $0<a<1$, it is decreasing. For example,

$$\log_{1/2}x=-\log_2x,$$

so $y=\log_{1/2}x$ is the reflection of $y=\log_2x$ in the $x$-axis.

Worked example 7: transform a logarithmic graph

Describe the graph

$$y=\ln(x-3)+2.$$

Starting from $y=\ln x$:

• $x-3$ translates the graph $3$ units right;
• $+2$ translates it $2$ units up.

The asymptote $x=0$ becomes $x=3$, and the domain becomes

$$x>3.$$

For the $x$-intercept, set $y=0$:

$$\ln(x-3)+2=0 \quad\Longrightarrow\quad \ln(x-3)=-2$$ $$\Longrightarrow\quad x-3=e^{-2} \quad\Longrightarrow\quad x=3+e^{-2}.$$

Thus the exact intercept is $(3+e^{-2},0)$. There is no $y$-intercept because $x=0$ is outside the domain.

Common misconceptions

• A logarithm is not a base times an argument. $\log_28$ means the exponent $3$, not $2\times8$.
• Arguments must be positive. In $\ln(x-4)$, require $x-4>0$, not merely $x>0$.
• Products split, sums do not. $\ln(xy)=\ln x+\ln y$, but $\ln(x+y)$ cannot be split.
• A power becomes a coefficient. $\ln(x^3)=3\ln x$, whereas $(\ln x)^3$ is a different expression.
• Use one base consistently in change of base. Both numerator and denominator must use the same logarithm function.
• Do not round midway. Keep $\ln20/\ln7$ in the calculator and round only the final result.

Mixed self check

1. State the domain of $f(x)=\log_3(7-2x)$.
2. Simplify $\log_a(a^5)+\log_a(1/a^2)$.
3. Write $\ln12+\ln x-\ln4$ as one logarithm.
4. Solve $\log_5x=-2$.
5. Evaluate $\log_49$ to three significant figures.
6. Explain why $\ln(x^2)=2\ln x$ is not valid for every nonzero real $x$.

Answers

1. Require $7-2x>0$, so $x<7/2$.
2. $5+(-2)=3$.
3. $\ln(3x)$, with $x>0$.
4. $x=5^{-2}=1/25$.
5. $\displaystyle \log_49=\frac{\ln9}{\ln4}=1.58$ to three significant figures.
6. If $x<0$, then $\ln x$ is not real. The valid identity for every $x\ne0$ is $\ln(x^2)=2\ln|x|$. At this stage, it is safest to state $x>0$ before using $\ln(x^2)=2\ln x$.

What to learn next

You should now be able to translate between logarithmic and exponential form, apply the three laws in either direction, change base and identify the domain of a logarithm. Next, use these skills to solve exponential and logarithmic equations. They also underpin growth and decay models and linearising data.