F4, F5 - Logarithm laws and equations | AQA A-Level Mathematics

Use F4 logarithm laws to combine expressions and check their domains, then use F5 to solve equations with an unknown exponent. Exact logarithmic answers and checks in the original equation keep the algebra reliable.

Using logarithm laws

Can $\ln(x-1)+\ln(x+2)=\ln10$ have an algebraic root that is not a solution of the logarithmic equation?

We will return to that puzzle after practising the laws. Start small: $\log_2 4+\log_2 8=2+3=5=\log_2 32$ . Addition of logarithms multiplies the arguments. It does not add them.

You already know

$\log_a v=u$ means $a^u=v$ , where $a>0$ , $a\ne1$ and $v>0$ . In particular, $\log_a1=0$ , $\log_a a=1$ and $\log_a(a^r)=r$ . These results are not in the formula booklet: learn them.

Write $u=a^p$ and $v=a^q$ . Multiplication, division and powers give $a^{p+q}$ , $a^{p-q}$ and $a^{pr}$ ; taking logarithms returns those exponents. That is why all three laws share the same base and require positive arguments.

Product, quotient and power laws

For $u,v>0$ , $a>0$ , $a\ne1$ and real $r$ ,

\begin{aligned} \log_a(uv)&=\log_a u+\log_a v \\ \log_a(u/v)&=\log_a u-\log_a v \\ \log_a(u^r)&=r\log_a u \end{aligned}

These include $\log_a(1/u)=-\log_a u$ and $\log_a(1/\sqrt u)=-\frac12\log_a u$ . They also hold for $\ln$ .

Not in the formula booklet: learn it.

Negative and fractional powers belong to the same law. A reciprocal contributes a minus sign; a square root contributes a factor $1/2$ . Keep brackets around a denominator until all its factors have been expanded.

Worked example 1: a root in the denominator

Question. For $u,v>0$ , expand $\ln(12/(u\sqrt v))$ .

Plan. Separate the quotient, then the denominator's product, then its square root.

\begin{aligned} \ln\left(\frac{12}{u\sqrt v}\right)&=\ln12-\ln(u\sqrt v) && \text{quotient law} \\ &=\ln12-(\ln u+\ln\sqrt v) && \text{product law} \\ &=\ln12-\ln u-\tfrac12\ln v && \text{power law} \end{aligned}

Why this step? Subtracting the whole denominator log means that both its expanded terms must be subtracted.

Answer. $\ln12-\ln u-\frac12\ln v$ .

Check. At $u=3$ , $v=4$ , the original is $\ln2$ ; the expansion is $\ln12-\ln3-\ln2=\ln2$ .

Change only the power of $u$ from $1$ to $2$ . The corresponding coefficient doubles; the treatment of the root is unchanged.

Worked example 2: a squared denominator factor

Question. For $u,v>0$ , expand $\ln(12/(u^2\sqrt v))$ .

Plan. Use the same quotient and product steps, then apply the power law to both variable factors.

\begin{aligned} \ln\left(\frac{12}{u^2\sqrt v}\right)&=\ln12-\ln(u^2\sqrt v) && \text{quotient law} \\ &=\ln12-\ln(u^2)-\ln(v^{1/2}) && \text{product law} \\ &=\ln12-2\ln u-\tfrac12\ln v && \text{power law} \end{aligned}

Why this step? Both powers are inside a subtracted logarithm, so both coefficients remain negative.

Answer. $\ln12-2\ln u-\frac12\ln v$ .

Check. At $u=2$ , $v=9$ , the original argument is $12/(4\times3)=1$ . The expanded logs also combine to $\ln1=0$ .

The laws work backwards too. For $x,y>0$ , $2\ln x-\frac12\ln y-\ln3=\ln(x^2/(3\sqrt y))$ . A number outside a log must first be written in the same base: if $\log_3 z=2+\log_3 5$ , then $2=\log_3 9$ , so $z=9\times5=45$ . Both are consequences of the unsupplied laws above.

Watch out

Wrong: $\ln12-\ln(u\sqrt v)=\ln12-\ln u+\frac12\ln v$ . The minus sign applies to the whole expansion of the denominator. The last term must be $-\frac12\ln v$ .

Change only the numerator constant in example 2. Keep the restrictions as you expand.

Solving exponential equations

A logarithm brings an unknown exponent down into ordinary algebra.

For example, $2^x=8$ gives $x=3$ ; when the right-hand side is $7$ , the exact exponent still exists even though it is not an integer. Taking the same logarithm on both sides gives $x\ln2=\ln7$ .

Solving a single exponential

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

a^x=b\quad\Longrightarrow\quad x=\frac{\ln b}{\ln a}=\log_a b

Any one valid log base may be used on both sides. For $e^x=b$ , this reduces to $x=\ln b$ .

Not in the formula booklet: learn it.

First isolate the exponential. Thus $4e^{-0.3t}=1.5$ becomes $e^{-0.3t}=3/8$ , then $-0.3t=\ln(3/8)$ , giving $t=\ln(8/3)/0.3\approx3.269$ . An isolated exponential equal to zero or a negative number has no real solution. A base of $1$ is exceptional: $1^x=b$ holds for every real $x$ if $b=1$ , and for none otherwise.

Worked example 3: a complete exponent

Question. Solve $7^{2x-1}=20$ , giving an exact answer and a value to three decimal places.

Plan. Take $\ln$ of both sides, use the power law on the whole exponent, then isolate $x$ .

\begin{aligned} (2x-1)\ln7&=\ln20 && \text{same logarithm on both sides} \\ 2x-1&=\frac{\ln20}{\ln7} && \text{divide by nonzero }\ln7 \\ x&=\frac12\left(1+\frac{\ln20}{\ln7}\right) && \text{solve the linear equation} \end{aligned}

Why this step? The power law multiplies $\ln7$ by all of $2x-1$ , not just by $2x$ .

Answer. $x=\frac12(1+\ln20/\ln7)\approx1.270$ .

Check. The unrounded value gives $2x-1=\ln20/\ln7$ , so $7^{2x-1}=e^{\ln20}=20$ .

Change only the right-hand side to its reciprocal. This reverses the sign of the log value, but does not negate the entire answer for $x$ : the exponent also contains $-1$ .

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F4, F5 - Logarithm laws and equations

Using logarithm laws

Solving exponential equations

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