Solving Exponential and Log Equations

Solve equations with x in the power by taking logs, spot hidden quadratics in eˣ, 2ˣ or ln x, and solve log equations, rejecting impossible roots.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 41 real past-paper questions below

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The unknown in the power

Money in a bank grows by 5% every year. When will it double? The unknown is the number of years, and it sits in the power. Logs bring it down.

If both sides are powers of the same base, like $2^{x} = 32 = 2^{5}$ , just match the powers. When they are not, take logs of both sides.

$a^{x} = b$ — the unknown is in the power.

$l g (a^{x}) = l g b$ — take lg (log to base 10) of both sides. Any base works, but your calculator has lg and ln.

$x l g a = l g b$ — power law: the power comes down to the front.

$x = \frac{l g b}{l g a}$ — divide both sides by lg a.

Formula

x = \frac{l g b}{l g a}

$b$ — the number on the other side of the equation.
$a$ — the base that has x in its power.

Using ln instead, $x = \frac{l n b}{l n a}$ , gives the same answer. No board prints this method, so learn the steps.

Worked example

1
Solve $3^{x} = 20$ , giving x to 3 significant figures.
2
$x l g 3 = l g 20$ — Take lg of both sides. The power law brings x down.
3
$x = \frac{l g 20}{l g 3}$ — Divide by lg 3.
4
$x = 2.73$ — 2.7268..., rounded to 3 significant figures.
5
$3^{2.73} \approx 20$ — Check with your calculator: it works.

Worked example

1
Solve $e^{x} = 7$ , giving x exactly.
2
$x = ln 7$ — Take ln of both sides. ln undoes e.

Rule

No calculator on 0606 Paper 1: leave answers exact, such as $x = ln 7$ or $x = \frac{l g 20}{l g 3}$ . On a calculator paper, give 3 significant figures unless the question says otherwise.

Common mistake

$\frac{l g 20}{l g 3}$ is NOT $l g (\frac{20}{3})$ . A fraction of two logs is not the log of a fraction.

Key idea

Radiocarbon dating finds the age of old bones and wood by solving an exponential equation like these. Willard Libby developed the method and won the 1960 Nobel Prize in Chemistry for it.

Key idea

Take logs of both sides. The power law brings x down to the ground.

Brackets and two bases

Sometimes the power is a bracket, or there are powers on both sides. The plan is the same: take logs, bring the powers down, then solve the ordinary equation that is left.

Worked example

1
Solve $2^{x + 1} = 5^{x}$ , giving x to 3 significant figures.
2
$(x + 1) l g 2 = x l g 5$ — Take lg of both sides. The whole bracket comes down as the power.
3
$x l g 2 + l g 2 = x l g 5$ — Expand the bracket.
4
$l g 2 = x (l g 5 - l g 2)$ — Collect the x terms on one side and factorise.
5
$x = \frac{l g 2}{l g 5 - l g 2}$ — Divide by the bracket.
6
$x = 0.756$ — 0.75647..., rounded to 3 significant figures.

Worked example

1
Solve $e^{3 x} = 6 e^{x + 2}$ , giving x exactly.
2
$\frac{e ^{3 x}}{e ^{x + 2}} = 6$ — Get the e terms together: divide both sides by $e^{x + 2}$ .
3
$e^{2 x - 2} = 6$ — Dividing powers: subtract the indices. $3 x - (x + 2) = 2 x - 2$ .
4
$2 x - 2 = ln 6$ — Take ln of both sides.
5
$x = 1 + \frac{1}{2} ln 6$ — Add 2, then halve.

Common mistake

Writing $l g 2^{x + 1}$ as $x + 1 l g 2$ . The whole power comes down in a bracket: $(x + 1) l g 2$ .

Key idea

Logs turn a power equation into an ordinary linear equation. Then solve it the way you always do.

Least n: the sign flip

Some questions ask for the least whole number n. For example, a car loses 20% of its value every year, so after n years it keeps $0. 8^{n}$ of its price.

Worked example

1
Find the least whole number of years, n, after which the car is worth less than 10% of its price.
2
$0. 8^{n} < 0.1$ — Less than 10% means less than 0.1 of the price.
3
$n l g 0.8 < l g 0.1$ — Take lg of both sides. The power law brings n down.
4
$n > \frac{l g 0.1}{l g 0.8}$ — lg 0.8 is negative. Dividing by a negative number turns the sign round.
5
$n > 10.3$ — 10.318... to 3 significant figures.
6
$n = 11$ — n must be a whole number bigger than 10.3.
7
$0. 8^{10} \approx 0.107, 0. 8^{11} \approx 0.086$ — Check: 10 years is not enough, 11 years is.

Rule

The log of a number between 0 and 1 is negative. When you divide both sides of an inequality by a negative number, turn the sign round.

Common mistake

Rounding 10.3 down to 10. The question wants the first whole number that works, so round up to 11.

Key idea

Check your n and the number just below it. That proves you have the least value.

Hidden quadratics in powers

Look at $2^{2 x} - 6 \times 2^{x} + 8 = 0$ . It is a quadratic in disguise, because $2^{2 x}$ is the square of $2^{x}$ . Call $2^{x}$ a new letter, u, and the disguise falls off.

Formula

2^{2 x} = (2^{x})^{2}

$2^{x}$ — the hidden unknown. Let $u = 2^{x}$ , so $2^{2 x} = u^{2}$ .

Three more index rewrites you will need, all from the laws of indices:

$2^{x + 1} = 2 \times 2^{x} = 2 u$
$2^{2 x + 1} = 2 \times 2^{2 x} = 2 u^{2}$
$e^{- x} = \frac{1}{e ^{x}}$ , so multiply through by $e^{x}$ to clear it.

Worked example

1
Solve $2^{2 x + 1} - 9 \times 2^{x} + 4 = 0$ .
2
$2 u^{2} - 9 u + 4 = 0$ — Let $u = 2^{x}$ . Then $2^{2 x + 1} = 2 u^{2}$ .
3
$(2 u - 1) (u - 4) = 0$ — Factorise.
4
$u = \frac{1}{2} or u = 4$ — Each bracket equals 0.
5
$2^{x} = \frac{1}{2} \Rightarrow x = - 1$ — Put $2^{x}$ back. $2^{- 1} = \frac{1}{2}$ .
6
$2^{x} = 4 \Rightarrow x = 2$ — $2^{2} = 4$ .

Worked example

1
Solve $2 e^{x} - 3 e^{- x} = 5$ , giving x exactly.
2
$2 e^{2 x} - 3 = 5 e^{x}$ — Multiply every term by $e^{x}$ . Note $e^{- x} \times e^{x} = 1$ .
3
$2 u^{2} - 5 u - 3 = 0$ — Let $u = e^{x}$ and move everything to one side.
4
$(2 u + 1) (u - 3) = 0$ — Factorise.
5
$e^{x} = - \frac{1}{2}$ — Impossible: $e^{x}$ is always positive. Reject it.
6
$e^{x} = 3 \Rightarrow x = ln 3$ — Take ln of both sides.

Rule

Mark schemes give a mark for the 3-term quadratic in $2^{x}$ or $e^{x}$ , then a mark for solving it. Write the substitution line so the examiner can see it.

Common mistake

Keeping an answer like $e^{x} = - \frac{1}{2}$ . A positive base to any power is positive, so reject it.

Key idea

Spot the square: $2^{2 x}$ next to $2^{x}$ , or $e^{- x}$ next to $e^{x}$ , means a hidden quadratic.

Hidden quadratics in logs

The same trick works with logs. If you see $(ln x)^{2}$ next to $ln x$ , let $u = ln x$ .

Worked example

1
Solve $2 (ln x)^{2} - 5 ln x + 2 = 0$ , giving x exactly.
2
$2 u^{2} - 5 u + 2 = 0$ — Let $u = ln x$ .
3
$(2 u - 1) (u - 2) = 0$ — Factorise.
4
$ln x = \frac{1}{2} \Rightarrow x = e^{\frac{1}{2}}$ — Raise e to both sides. e undoes ln.
5
$ln x = 2 \Rightarrow x = e^{2}$ — The same again.

Some questions hide the quadratic using two bases, like $lo g_{3} x$ and $lo g_{x} 3$ . The swap rule from the laws lesson joins them: $lo g_{x} 3 = \frac{1}{l o g _{3} x}$ .

Worked example

1
Solve $lo g_{3} x + 2 lo g_{x} 3 = 3$ .
2
$u + \frac{2}{u} = 3$ — Let $u = lo g_{3} x$ . By the swap rule, $lo g_{x} 3 = \frac{1}{u}$ .
3
$u^{2} - 3 u + 2 = 0$ — Multiply every term by u, then move everything to one side.
4
$(u - 1) (u - 2) = 0$ — Factorise.
5
$lo g_{3} x = 1 \Rightarrow x = 3$ — Index form: $x = 3^{1}$ .
6
$lo g_{3} x = 2 \Rightarrow x = 9$ — Index form: $x = 3^{2}$ .

Rule

Mark schemes give the first mark for a correct change to one base. Show that line before you form the quadratic.

Common mistake

Stopping at $lo g_{3} x = 2$ . The question asks for x, so turn each log back into index form.

Key idea

Change to one base, let u be the log, solve the quadratic, then undo the log.

Log equations

Many equations have logs on both sides, or a log equal to a number. Combine the logs into one single log, then remove it.

Formula

lo g_{a} P = k ⟺ P = a^{k}

$P$ — the inside of one single log.
$k$ — a plain number. It becomes the power on the base.

Formula

lo g_{a} P = lo g_{a} Q ⟺ P = Q

$P$ — the inside of the single log on the left.
$Q$ — the inside of the single log on the right, with the same base.

Worked example

1
Solve $lo g_{2} x + lo g_{2} (x - 2) = 3$ .
2
$lo g_{2} (x (x - 2)) = 3$ — Product law: one single log.
3
$x (x - 2) = 2^{3} = 8$ — Remove the log: write it in index form.
4
$x^{2} - 2 x - 8 = 0$ — Expand and move everything to one side.
5
$(x - 4) (x + 2) = 0$ — Factorise.
6
$x = 4$ — Reject $x = - 2$ : $lo g_{2} (- 2)$ does not exist.

Worked example

1
Solve $lo g_{5} (x + 4) - lo g_{25} x = 1$ .
2
$lo g_{25} x = \frac{l o g _{5} x}{l o g _{5} 25} = \frac{1}{2} lo g_{5} x$ — Change to base 5. $lo g_{5} 25 = 2$ .
3
$2 lo g_{5} (x + 4) - lo g_{5} x = 2$ — Multiply every term by 2.
4
$lo g_{5} \frac{( x + 4 ) ^{2}}{x} = 2$ — Power law, then quotient law.
5
$(x + 4)^{2} = 25 x$ — Remove the log: $5^{2} = 25$ . Then multiply by x.
6
$x^{2} - 17 x + 16 = 0$ — Expand: $x^{2} + 8 x + 16 = 25 x$ .
7
$x = 1 or x = 16$ — $(x - 1) (x - 16) = 0$ . Both keep every inside positive, so keep both.

Rule

Check each answer in the ORIGINAL equation. The inside of every log must be positive. Mark schemes expect an invalid root to be rejected.

Common mistake

Removing logs term by term. $lo g_{2} x + lo g_{2} (x - 2) = 3$ does NOT become $x + (x - 2) = 3$ . Combine into one log first.

Key idea

One base, one single log, then remove it. Finally, check that every inside is positive.

Which method? Read the question

Before you start, read what the question gives you. Its shape tells you which method to use.

You see	You do
$a^{x} = b$	Take lg or ln of both sides
Powers on both sides	Take logs, expand, collect the x terms
“Least whole number n”	Take logs; turn the sign if you divide by a negative log; round up
$2^{2 x}$ and $2^{x}$ , or $e^{x}$ and $e^{- x}$	Let $u = 2^{x}$ or $u = e^{x}$ and solve a quadratic
$(ln x)^{2}$ , or $lo g_{a} x$ with $lo g_{x} a$	One base, let u be the log, solve a quadratic
Logs joined by + or −	Combine into one log, then remove it

Common mistake

Skipping the final check. Reject any answer that gives the log of 0 or of a negative number, or a power equal to a negative number.

Key idea

Every method ends the same way: undo the log or the power, then check.

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