Mathematics 9709 · Pure Mathematics 2

Solving ln / log equations (combine, then undo with e)

The safest mark in the chapter: a Q1 or Q2 on almost every P3 paper.
One fixed move: squash both sides to a single log, then undo the log with $e$ .
If it is log = log, just match what is inside instead.

How to spot it: one equation with $\ln$ (or $\log$ ) wrapped around $x$ -terms on one side or both. The move: combine to one log on each side. Then either match what is inside (log = log) or raise $e$ to both sides (log = number).

The fixed method

Use the laws to get one log on each side (or a single log equal to a number).
If $\ln A=\ln B$ , what is inside must match: $A=B$ (equality of logarithms).
If $\ln A=k$ (a plain number), undo the $\ln$ by raising $e$ to both sides: $A=e^k$ .
Solve the resulting linear or quadratic equation.
Reject any root that makes a log's inside $\le 0$ — logs only work on positive numbers.

Calculator-first still works for the final number. But steps 1–3 are the method marks. Write them down even when the calculator gives you the root. (textbook §2.4 method)

Two words in the question tell you which path to take. See “exact form” or “show that” → keep the right side a log, so match what is inside and leave surds. See “3 d.p.” or “3 s.f.” → the right side is a number, so take $e$ of both sides.

Log = number → take e of both sides

Worked example

Solve $\ln(x+5)=5+\ln x$ , to 3 d.p.

Get the logs together. Subtract $\ln x$ and use the quotient law: $\ln(x+5)-\ln x=5\Rightarrow\ln\!\frac{x+5}{x}=5$ .
Now it is log = number, so take $e$ of both sides: $\frac{x+5}{x}=e^5$ .
Clear the fraction (the $e^5$ multiplies all of $x$ ): $x+5=x\,e^5\Rightarrow 5=x(e^5-1)$ .
Isolate: $x=\dfrac{5}{e^5-1}=0.034$ .

Answer

x=\dfrac{5}{e^5-1}=0.034\ \text{(3 d.p.)}

(9709/33 May/Jun 2023 Q1)

Common mistake

The lost mark here is where the $e^5$ lands. The left side is

\frac{x+5}{x}

, so taking

e

gives

x+5=x\,e^5

. The

e^5

multiplies the

x

, not the

5

. Writing

x+5=e^5

forgets the

x

on the bottom, and that loses the mark.

When combining gives a quadratic

Sometimes squashing the logs leaves a quadratic inside, but the move is the same.

Worked example

Solve $\ln(x-5)=7-\ln x$ , to 2 d.p.

Pull the logs together with the product law: $\ln(x-5)+\ln x=7\Rightarrow\ln\!\big(x(x-5)\big)=7$ .
Take $e$ of both sides: $x(x-5)=e^7\Rightarrow x^2-5x-e^7=0$ .
Calculator-first on the quadratic gives $x=35.71$ or $x=-30.71$ .
Reject $x=-30.71$ : the original $\ln(x-5)$ needs $x>5$ .

Answer

x=35.71\ \text{(2 d.p.)}

(9709/31 May/Jun 2024 Q2)

Common mistake

Always put your roots back into the original equation. Examiners plant a negative root on purpose. Hand in both and you lose the accuracy mark.

Exact form: log = log, write the algebra

When the question says “exact form” or “show that”, a decimal scores zero, so leave the answer as a surd, fraction or $e^k$ .

Worked example

Solve $\ln(2x^2-3)=2\ln x-\ln 2$ , exact form

Build the right side into one log: $2\ln x-\ln 2=\ln x^2-\ln 2=\ln\!\frac{x^2}{2}$ .
Now $\ln$ equals $\ln$ , so what is inside must match: $2x^2-3=\dfrac{x^2}{2}$ .
Clear the fraction and collect: $4x^2-6=x^2\Rightarrow 3x^2=6\Rightarrow x^2=2$ .
Square-root: $x=\pm\sqrt2$ , but $\ln x$ on the right needs $x>0$ , so reject $-\sqrt2$ .

Answer

x=\sqrt2

(9709/32 May/Jun 2023 Q2). The mark scheme gives M0 if you write $2\ln x-\ln 2=\ln(2x)$ : the $2$ is a number out front, not part of the inside.

Worked example

Solve $2\ln(2x+3)-\ln(2x+5)=\ln 3x$ , exact form

A three-term version. Same move, just keep the algebra tidy.

Power then quotient on the left: $\ln(2x+3)^2-\ln(2x+5)=\ln\!\dfrac{(2x+3)^2}{2x+5}$ .
Now $\ln=\ln$ , so equate what is inside: $\dfrac{(2x+3)^2}{2x+5}=3x$ .
Clear and expand: $(2x+3)^2=3x(2x+5)\Rightarrow 4x^2+12x+9=6x^2+15x$ .
Collect: $2x^2+3x-9=0\Rightarrow(2x-3)(x+3)=0$ , so $x=\tfrac32$ or $x=-3$ .
Reject $x=-3$ : it makes $\ln 3x$ a log of a negative.

Answer

x=\tfrac32

(9709/33 May/Jun 2025 Q2)

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2022 Q1)

Solve $\ln(2x-1)=2\ln(x+1)-\ln x$ , to 3 d.p. Build the right side into one log: $\ln\!\frac{(x+1)^2}{x}$ . Match what is inside: $x(2x-1)=(x+1)^2$ . Expand to $x^2-3x-1=0$ , so $x=3.303$ (throw out the negative root, which fails $2x-1>0$ ).

Answer

x=3.303\ \text{(3 d.p.)}

Your turn— tap to reveal the worked answer (9709/32 Feb/Mar 2020 Q2)

Solve $\ln 3+\ln(2x+5)=2\ln(x+2)$ in exact form. Combine each side to one log: $\ln\!\big(3(2x+5)\big)=\ln(x+2)^2$ . Match what is inside: $3(2x+5)=(x+2)^2$ . Expand to $x^2-2x-11=0$ , giving $x=1\pm2\sqrt3$ . Throw out $1-2\sqrt3$ (it makes $x+2<0$ ).

Answer

x=1+2\sqrt3

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Solving ln / log equations (combine, then undo with e)

The safest mark in the chapter: a Q1 or Q2 on almost every P3 paper.
One fixed move: squash both sides to a single log, then undo the log with $e$ .
If it is log = log, just match what is inside instead.

The fixed method

Use the laws to get one log on each side (or a single log equal to a number).
If $\ln A=\ln B$ , what is inside must match: $A=B$ (equality of logarithms).
If $\ln A=k$ (a plain number), undo the $\ln$ by raising $e$ to both sides: $A=e^k$ .
Solve the resulting linear or quadratic equation.
Reject any root that makes a log's inside $\le 0$ — logs only work on positive numbers.

Calculator-first still works for the final number. But steps 1–3 are the method marks. Write them down even when the calculator gives you the root. (textbook §2.4 method)

Log = number → take e of both sides

Worked example

Solve $\ln(x+5)=5+\ln x$ , to 3 d.p.

Get the logs together. Subtract $\ln x$ and use the quotient law: $\ln(x+5)-\ln x=5\Rightarrow\ln\!\frac{x+5}{x}=5$ .
Now it is log = number, so take $e$ of both sides: $\frac{x+5}{x}=e^5$ .
Clear the fraction (the $e^5$ multiplies all of $x$ ): $x+5=x\,e^5\Rightarrow 5=x(e^5-1)$ .
Isolate: $x=\dfrac{5}{e^5-1}=0.034$ .

Answer

x=\dfrac{5}{e^5-1}=0.034\ \text{(3 d.p.)}

(9709/33 May/Jun 2023 Q1)

Common mistake

The lost mark here is where the $e^5$ lands. The left side is

\frac{x+5}{x}

, so taking

e

gives

x+5=x\,e^5

. The

e^5

multiplies the

x

, not the

5

. Writing

x+5=e^5

forgets the

x

on the bottom, and that loses the mark.

When combining gives a quadratic

Sometimes squashing the logs leaves a quadratic inside, but the move is the same.

Worked example

Solve $\ln(x-5)=7-\ln x$ , to 2 d.p.

Pull the logs together with the product law: $\ln(x-5)+\ln x=7\Rightarrow\ln\!\big(x(x-5)\big)=7$ .
Take $e$ of both sides: $x(x-5)=e^7\Rightarrow x^2-5x-e^7=0$ .
Calculator-first on the quadratic gives $x=35.71$ or $x=-30.71$ .
Reject $x=-30.71$ : the original $\ln(x-5)$ needs $x>5$ .

Answer

x=35.71\ \text{(2 d.p.)}

(9709/31 May/Jun 2024 Q2)

Common mistake

Always put your roots back into the original equation. Examiners plant a negative root on purpose. Hand in both and you lose the accuracy mark.

Exact form: log = log, write the algebra

When the question says “exact form” or “show that”, a decimal scores zero, so leave the answer as a surd, fraction or $e^k$ .

Worked example

Solve $\ln(2x^2-3)=2\ln x-\ln 2$ , exact form

Build the right side into one log: $2\ln x-\ln 2=\ln x^2-\ln 2=\ln\!\frac{x^2}{2}$ .
Now $\ln$ equals $\ln$ , so what is inside must match: $2x^2-3=\dfrac{x^2}{2}$ .
Clear the fraction and collect: $4x^2-6=x^2\Rightarrow 3x^2=6\Rightarrow x^2=2$ .
Square-root: $x=\pm\sqrt2$ , but $\ln x$ on the right needs $x>0$ , so reject $-\sqrt2$ .

Answer

x=\sqrt2

(9709/32 May/Jun 2023 Q2). The mark scheme gives M0 if you write $2\ln x-\ln 2=\ln(2x)$ : the $2$ is a number out front, not part of the inside.

Worked example

Solve $2\ln(2x+3)-\ln(2x+5)=\ln 3x$ , exact form

A three-term version. Same move, just keep the algebra tidy.

Power then quotient on the left: $\ln(2x+3)^2-\ln(2x+5)=\ln\!\dfrac{(2x+3)^2}{2x+5}$ .
Now $\ln=\ln$ , so equate what is inside: $\dfrac{(2x+3)^2}{2x+5}=3x$ .
Clear and expand: $(2x+3)^2=3x(2x+5)\Rightarrow 4x^2+12x+9=6x^2+15x$ .
Collect: $2x^2+3x-9=0\Rightarrow(2x-3)(x+3)=0$ , so $x=\tfrac32$ or $x=-3$ .
Reject $x=-3$ : it makes $\ln 3x$ a log of a negative.

Answer

x=\tfrac32

(9709/33 May/Jun 2025 Q2)

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2022 Q1)

Answer

x=3.303\ \text{(3 d.p.)}

Your turn— tap to reveal the worked answer (9709/32 Feb/Mar 2020 Q2)

Answer

x=1+2\sqrt3