P2 5.2 - Logarithms and their laws | Edexcel A-Level International Mathematics

Read a logarithm as a missing exponent, then use logarithm laws to simplify expressions and solve equations. Positive arguments matter throughout: you will learn why algebra can produce candidates that the original logarithms do not allow.

Logarithms as inverse powers

What power of $2$ gives $32$ , and what power gives $10$ ? The first answer is $5$ . The second lies between $3$ and $4$ , since $8<10<16$ . A logarithm gives that second exponent an exact name: $\log_2 10$ .

The subscript states the base. A logarithm is a number, namely the exponent needed to produce its argument.

Logarithm

For $a>0$ , $a\ne1$ and $b>0$ ,

\log_a b=c\quad\Longleftrightarrow\quad a^c=b.

The expression $\log_a b$ asks “to what power must $a$ be raised to obtain $b$ ?”

There is exactly one such real exponent: $a^x$ is strictly increasing or strictly decreasing, and takes every positive output once. The inverse relations are

a^{\log_a b}=b,\qquad \log_a(a^c)=c.

On many calculators, $\log$ without a written base means $\log_{10}$ . Keep a written base when it is different. Nothing in the meaning requires the answer to be positive or an integer.

Worked example 1: a positive logarithm

Question. Find $\log_3 81$ exactly.

Plan. Rewrite $81$ as a power of the stated base.

81=3^4\quad\Longrightarrow\quad\log_3 81=4.

Answer. $4$ .

Check. Reconstruct the argument: $3^4=81$ .

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

Worked example 2: a negative logarithm

Question. Find $\log_3(1/81)$ exactly.

Plan. Use the reciprocal rule for powers, then read off the exponent.

\frac1{81}=3^{-4}\quad\Longrightarrow\quad\log_3(1/81)=-4.

Answer. $-4$ .

Check. $3^{-4}=1/3^4=1/81$ ; the negative answer gives a positive argument below one.

The argument and the answer have different roles.

Watch out

A logarithm may be negative. Its argument must be positive. Thus $\log_3(1/81)$ exists, but $\log_3(-81)$ is not real.

Read the logarithm as a missing power.

Product quotient and power laws

Logarithm laws come from index laws: multiplying powers adds their exponents, so the logarithm of a product is a sum.

For a concrete case, $\log_2 8+\log_2 4=3+2=5$ , and $\log_2(8\times4)=\log_2 32=5$ . To see why this always works, put $x=a^p$ and $y=a^q$ . Then $xy=a^{p+q}$ , $x/y=a^{p-q}$ and $x^k=a^{pk}$ . Taking logarithms reads off the three exponents.

Product quotient and power laws

For $a>0$ , $a\ne1$ , $x>0$ , $y>0$ and real $k$ ,

\begin{aligned} \log_a(xy)&=\log_a x+\log_a y\\ \log_a\left(\frac{x}{y}\right)&=\log_a x-\log_a y\\ \log_a(x^k)&=k\log_a x. \end{aligned}

Not in the formula booklet: learn them.

You can use each law in either direction. To combine several logarithms, move any coefficient into its argument as a power first. All the logarithms being combined must have the same base.

Worked example 3: combining three logarithms

Question. Write $2\log_3 x+\log_3 5-\log_3 y$ as one logarithm, where $x,y>0$ .

Plan. Use the power law first, then turn addition into multiplication and subtraction into division.

\begin{aligned} 2\log_3 x+\log_3 5-\log_3 y &=\log_3(x^2)+\log_3 5-\log_3 y\\ &=\log_3\left(\frac{5x^2}{y}\right). \end{aligned}

Answer. $\log_3(5x^2/y)$ .

Check. At $x=3,y=5$ , the original is $2\times1+\log_3 5-\log_3 5=2$ . The combined argument is $9$ , and $\log_3 9=2$ .

Change only the coefficient of $\log_3 x$ from $2$ to $-2$ . The resulting negative power belongs in the denominator.

Worked example 4: a negative coefficient

Question. Write $-2\log_3 x+\log_3 5-\log_3 y$ as one logarithm, where $x,y>0$ .

Plan. Move $-2$ into the exponent, then simplify the reciprocal.

\begin{aligned} -2\log_3 x+\log_3 5-\log_3 y &=\log_3(x^{-2})+\log_3 5-\log_3 y\\ &=\log_3\left(\frac{5x^{-2}}{y}\right)\\ &=\log_3\left(\frac5{x^2y}\right). \end{aligned}

Answer. $\log_3(5/(x^2y))$ .

Check. At $x=3,y=5$ , the original is $-2$ . The combined expression is $\log_3(1/9)=-2$ .

Use brackets to make clear what is inside a logarithm.

Watch out

$\log_a x-\log_a y=\log_a(x/y)$ . It is not $(\log_a x)/(\log_a y)$ . Also, there is no sum law for $\log_a(x+y)$ : for example, $\log_2(4+4)=3$ , while $\log_2 4+\log_2 4=4$ .

Combine a new expression in the same order.

Reciprocal and base identities

A base raised to zero is one, and a base raised to one is itself; these two facts give useful logarithms immediately.

For example, $5^0=1$ gives $\log_5 1=0$ , and $5^1=5$ gives $\log_5 5=1$ . A reciprocal argument reverses the sign because $1/x=x^{-1}$ .

Base and reciprocal identities

For $a>0$ , $a\ne1$ and $x>0$ ,

\log_a1=0,\qquad\log_a a=1,\qquad \log_a\left(\frac1x\right)=-\log_a x.

These are consequences of the inverse-power meaning and the logarithm laws; learn them.

The reciprocal identity also follows from the quotient law: $\log_a(1/x)=\log_a1-\log_a x=0-\log_a x$ . For positive $x$ , both sides are defined. A reciprocal of the argument is different from a reciprocal of the whole logarithm.

Worked example 5: using identities without a calculator

Question. Simplify $\log_a(a^3)+\log_a(1/a)$ , where $a>0$ , $a\ne1$ .

Plan. Read off the first exponent and use the reciprocal identity in the second term.

\log_a(a^3)+\log_a(1/a)=3-\log_a a=3-1=2.

Answer. $2$ .

Check. Combine the arguments instead: $\log_a(a^3/a)=\log_a(a^2)=2$ .

Change only the exponent in the first argument from $3$ to $-3$ . The second term still equals $-1$ .

Worked example 6: two negative contributions

Question. Simplify $\log_a(a^{-3})+\log_a(1/a)$ , where $a>0$ , $a\ne1$ .

Plan. Apply the inverse identities before adding.

\log_a(a^{-3})+\log_a(1/a)=-3-1=-4.

Answer. $-4$ .

Check. The product of the arguments is $a^{-3}a^{-1}=a^{-4}$ , whose base- $a$ logarithm is $-4$ .

These identities hold for a decay base too: $\log_{1/2}(1/2)=1$ .

Watch out

$\log_a(1/x)$ means a logarithm of a reciprocal. It equals $-\log_a x$ , not $1/\log_a x$ . For instance, $\log_2(1/8)=-3$ , while $1/\log_2 8=1/3$ .

Use the same identities with different exponents.

Domain restrictions

Every argument in the original equation must be positive, even if a later algebraic equation permits more values.

For $\log_3(x-1)+\log_3(x+1)$ , the restrictions are $x-1>0$ and $x+1>0$ , giving $x>1$ . Combining the logarithms gives $\log_3(x^2-1)$ , which also exists for $x<-1$ . On that extra interval the combined expression cannot replace the two original logarithms: neither original argument is positive there.

Logarithmic domain

The domain of a real logarithmic expression contains only inputs for which every logarithm has a positive argument. Its base must also be positive and different from one. Record these restrictions before solving, then check every candidate in the original expression.

A logarithm of zero is not real either: $a^c$ never equals zero. When two logarithms with the same valid base are equal, their positive arguments are equal, because an exponential function takes each positive value once.

Worked example 7: rejecting an algebraic candidate

Question. Solve $\log_3(x-1)+\log_3(x+1)=1$ , showing all stages of working.

Plan. Establish the domain, combine the logarithms, then use their inverse-power meaning.

The original domain is $x>1$ . Within that domain,

\begin{aligned} \log_3((x-1)(x+1))&=1 &&\text{product law}\\ x^2-1&=3^1 &&\text{inverse power}\\ x^2&=4\\ x&=2\text{ or }-2. \end{aligned}

Answer. $x=2$ only; $-2$ fails $x>1$ .

Check. At $x=2$ , the original left side is $\log_3 1+\log_3 3=0+1=1$ . At $x=-2$ , the arguments would be $-3$ and $-1$ , so this candidate cannot be substituted into real logarithms.

Change only the right-hand side from $1$ to $2$ . The algebra changes, but the original domain does not.

Worked example 8: an exact surd solution

Question. Solve $\log_3(x-1)+\log_3(x+1)=2$ , showing all stages of working.

Plan. Retain $x>1$ , then combine and convert to a power equation.

\begin{aligned} \log_3(x^2-1)&=2\\ x^2-1&=9\\ x^2&=10\\ x&=\pm\sqrt{10}. \end{aligned}

Answer. $x=\sqrt{10}$ only, since $-\sqrt{10}<1$ .

Check. The retained root is greater than one, so both original arguments are positive. Their product is $(\sqrt{10}-1)(\sqrt{10}+1)=9$ , and their logarithms therefore sum to $\log_3 9=2$ .

The opening missing exponent is now named exactly: $\log_2 10$ . Logarithm laws manipulate such numbers without replacing them prematurely by rounded decimals.

Exam tip

When asked to show all stages, show the logarithm law and the algebra used to solve the resulting equation. Reject any candidate that makes an original argument zero or negative.

Only the constant on the right changes again.

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P2 5.2 - Logarithms and their laws

Logarithms as inverse powers

Product quotient and power laws

Reciprocal and base identities

Domain restrictions

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