Solving Exponential Equations

This lesson focuses on the techniques for solving equations where the unknown appears in the exponent. These questions feature heavily in A-Level papers and draw on your knowledge of index laws and logarithm laws.

Type 1: Equations Reducible to a Common Base

If both sides can be written as powers of the same base, equate the indices.

Example 1: Solve 4ˣ = 32

Write both sides as powers of 2: (2²)ˣ = 2⁵
2²ˣ = 2⁵
2x = 5
x = 5/2

Example 2: Solve 9^(2x-1) = 27ˣ

Write as powers of 3: (3²)^(2x-1) = (3³)ˣ
3^(4x-2) = 3^(3x)
4x - 2 = 3x
x = 2

Exam Tip: Always look for a common base first. This avoids logarithms entirely and gives exact answers.

Type 2: Taking Logarithms of Both Sides

When the bases cannot be matched, take logs (base 10 or natural log) of both sides.

Example: Solve 3ˣ = 7

Take log₁₀ of both sides: log(3ˣ) = log(7)
x × log(3) = log(7) [Power Law]
x = log(7) / log(3) = 0.8451.../0.4771... ≈ 1.771

Example: Solve 5^(2x+1) = 4ˣ

Take ln of both sides: (2x + 1) × ln(5) = x × ln(4)
2x × ln(5) + ln(5) = x × ln(4)
2x × ln(5) - x × ln(4) = -ln(5)
x(2ln(5) - ln(4)) = -ln(5)
x = -ln(5) / (2ln(5) - ln(4))
x ≈ -0.5887

Type 3: Hidden Quadratics

Some exponential equations can be transformed into quadratic equations using a substitution.

Example: Solve 4ˣ - 6 × 2ˣ + 8 = 0

Step 1 — Notice that 4ˣ = (2²)ˣ = (2ˣ)². Let y = 2ˣ:

y² - 6y + 8 = 0

Step 2 — Factorise:

(y - 2)(y - 4) = 0
y = 2 or y = 4

Step 3 — Substitute back:

2ˣ = 2, so x = 1
2ˣ = 4 = 2², so x = 2

Example: Solve 9ˣ - 4 × 3ˣ + 3 = 0

Let y = 3ˣ (since 9ˣ = (3ˣ)²):

y² - 4y + 3 = 0
(y - 1)(y - 3) = 0
y = 1 or y = 3
3ˣ = 1 gives x = 0
3ˣ = 3 gives x = 1

Exam Tip: When you see terms like 4ˣ and 2ˣ in the same equation, or 9ˣ and 3ˣ, always try the substitution y = (smaller base)ˣ to form a quadratic.

Type 4: Equations Involving eˣ

The same techniques apply. Use ln to undo eˣ.

Example: Solve e²ˣ - 5eˣ + 6 = 0

Let y = eˣ:

y² - 5y + 6 = 0
(y - 2)(y - 3) = 0
y = 2 or y = 3
eˣ = 2 gives x = ln(2) ≈ 0.693
eˣ = 3 gives x = ln(3) ≈ 1.099

Example: Solve 2e²ˣ + eˣ - 3 = 0

Let y = eˣ:

2y² + y - 3 = 0
(2y + 3)(y - 1) = 0
y = -3/2 or y = 1
eˣ = -3/2 has no solution (eˣ is always positive)
eˣ = 1 gives x = 0

Type 5: Equations Mixing Exponentials and Constants

Example: Solve 3ˣ⁺¹ = 2ˣ + 5

This cannot be solved algebraically in a neat closed form. In the exam, you would typically be asked to show graphically or numerically that a solution exists in a given interval, then use iteration. However, if the question says "solve" and gives a specific form, look for substitution or common base opportunities.

Checking Your Solutions

Always verify solutions by substituting back:

For x = 2 in 4ˣ = 32: is this wrong? 4² = 16 ≠ 32. But x = 5/2: 4^(5/2) = (√4)⁵ = 2⁵ = 32. Correct.
For hidden quadratics, remember that y = aˣ > 0, so reject negative values of y.

Summary

Common base: rewrite both sides with the same base and equate indices.
Take logs: when bases differ, take log₁₀ or ln of both sides and use the power law.
Hidden quadratic: substitute y = aˣ when you spot terms like (a²)ˣ and aˣ together.
Always reject negative solutions for y = aˣ since exponential functions are always positive.
Verify your answer by substituting back into the original equation.

A-Level Deep Dive: Solving Exponential Equations

Spec mapping

Edexcel 9MA0-01 specification section 6 — Exponentials and logarithms, sub-strands 6.4 and 6.5 covers logarithms to solve equations and inequalities in which the unknown appears in the exponent (refer to the official specification document for exact wording). and "use logarithmic graphs to estimate parameters in relationships of the form $y = ax^n$ and $y = kb^x$ ." This sub-strand sits at the synoptic centre of Paper 1: every exponential-modelling question, every "decay constant" Mechanics problem, and every "find the time at which …" applied question funnels back through the procedure of taking a logarithm of both sides. Synoptic threads radiating outward include section 2 (Algebra and functions — index laws $a^{m+n} = a^m a^n$ and substitution to reduce to a quadratic), section 6.2-6.3 (log laws $\log(xy) = \log x + \log y$ and the change-of-base relation), the substitution $u = e^x$ that converts disguised exponential equations into routine quadratics, modelling questions on Paper 2 where exponential growth/decay couple back to logarithms for inversion, and numerical methods (section 10) for transcendental equations like $e^x = 5x$ that resist closed-form attack. The Edexcel formula booklet provides $\log_a x = \frac{\log_b x}{\log_b a}$ but does not list the index laws or the basic identity $\ln e^x = x$ — these must be memorised.

Worked example with full mark scheme

Question (8 marks):

(a) Solve the equation $5e^{2x} - 3e^x - 2 = 0$ , giving your answers as exact values where appropriate. (6)

(b) Hence, or otherwise, solve $5 \cdot 4^{2y} - 3 \cdot 4^y - 2 = 0$ , giving $y$ to 3 significant figures. (2)

Solution with mark scheme:

(a) Step 1 — recognise the hidden quadratic.

Notice that $e^{2x} = (e^x)^2$ . Let $u = e^x$ , so $u > 0$ for all real $x$ . The equation becomes:

$5u^2 - 3u - 2 = 0$

M1 — substitution $u = e^x$ (or equivalent), reducing the equation to a quadratic in $u$ . Examiners look for explicit declaration of the substitution; merely writing $5(e^x)^2 - 3e^x - 2 = 0$ without naming $u$ is acceptable but riskier on later steps where the bookkeeping matters.

Step 2 — factorise.

$5u^2 - 3u - 2 = (5u + 2)(u - 1) = 0$

So $u = -\dfrac{2}{5}$ or $u = 1$ .

M1 — correct factorisation (or use of the quadratic formula). Candidates who use the formula must show $u = \dfrac{3 \pm \sqrt{9 + 40}}{10} = \dfrac{3 \pm 7}{10}$ to bank the M1.

A1 — both roots correctly identified.

Step 3 — reject the negative root.

Since $u = e^x > 0$ for all real $x$ , we must reject $u = -\tfrac{2}{5}$ . The phrase examiners reward is "since $e^x > 0$ , we reject $u = -\tfrac{2}{5}$ ".

B1 — explicit rejection with justification. This mark is very commonly lost: candidates spot the negative root, silently discard it, and lose B1 because the rejection isn't articulated. The mark scheme treats the unspoken rejection as ambiguous between "knew to reject" and "got lucky".

Step 4 — solve the surviving exponential.

$e^x = 1 \implies x = \ln 1 = 0$ .

M1 — taking $\ln$ of both sides (or recognising $e^0 = 1$ ).

A1 — $x = 0$ .

(b) Step 1 — recognise the same structure.

Setting $v = 4^y$ , the equation $5v^2 - 3v - 2 = 0$ has the same factorisation as part (a): $v = 1$ (rejecting $v = -\tfrac{2}{5}$ since $4^y > 0$ ).

M1 — explicit reuse of the part (a) factorisation. The "Hence" command demands this connection; restarting from scratch costs nothing in marks here but signals a candidate who has missed the synoptic prompt.

Step 2 — invert to find $y$ .

$4^y = 1 \implies y \ln 4 = \ln 1 = 0 \implies y = 0$ (exact).

A1 — $y = 0$ , with the "to 3 s.f." instruction satisfied trivially because the answer is exact zero. A subtle examiner-craft point: writing $0.000$ to honour the 3 s.f. request shows precision, but $0$ is universally accepted.

Total: 8 marks (M4 A3 B1, split as shown).

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A radioactive isotope decays according to $N = N_0 e^{-kt}$ where $N$ is the number of nuclei present at time $t$ years, $N_0$ is the initial number, and $k > 0$ is a constant.

(a) Given that the half-life is 24 years, find $k$ in exact form. (3)

(b) Find, to the nearest year, the time taken for the sample to decay to 5% of its initial value. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO3.1a) — translating "half-life" into the equation $\tfrac{1}{2}N_0 = N_0 e^{-24k}$ , cancelling $N_0$ to give $e^{-24k} = \tfrac{1}{2}$ .
M1 (AO1.1b) — taking $\ln$ of both sides: $-24k = \ln(\tfrac{1}{2}) = -\ln 2$ .
A1 (AO1.1b) — $k = \dfrac{\ln 2}{24}$ in exact form.

(b)

M1 (AO3.1b) — setting $0.05 N_0 = N_0 e^{-kt}$ and cancelling $N_0$ .
M1 (AO1.1b) — taking $\ln$ : $-kt = \ln(0.05)$ , so $t = -\dfrac{\ln(0.05)}{k} = \dfrac{24 \ln 20}{\ln 2}$ using part (a).
A1 (AO1.1b) — $t \approx 104$ years (to nearest year).

Total: 6 marks split AO1 = 4, AO3 = 2. This is a classic AO3-flavoured modelling question — the maths is procedurally straightforward, but the marks reward translating the verbal description into algebra cleanly. Candidates who set up $N = \tfrac{1}{2}$ rather than $\tfrac{1}{2}N_0 = N_0 e^{-24k}$ collapse $N_0$ prematurely and lose AO3 marks.

Synoptic links

Connects to:

Section 2 — Algebra and functions (index laws and the substitution method): the substitution $u = e^x$ that converts $5e^{2x} - 3e^x - 2 = 0$ into a quadratic is the same technique used for $2^{2x} - 5 \cdot 2^x + 4 = 0$ and for surd equations $x - 5\sqrt{x} + 6 = 0$ (substitute $u = \sqrt{x}$ ). Recognising "term in $a^{2x}$ + term in $a^x$ + constant" as a hidden quadratic is the unifying synoptic insight.
Section 6.2-6.3 — Log laws and change of base: solving $3^x = 7$ requires $x = \log_3 7 = \dfrac{\ln 7}{\ln 3}$ via change of base. Without confidence in $\log(xy) = \log x + \log y$ and $\log(x^n) = n \log x$ , even the most basic exponential equation becomes inaccessible.
Section 7 — Differentiation of $e^x$ and $\ln x$ : the chain-rule derivative $\dfrac{d}{dx}e^{f(x)} = f'(x) e^{f(x)}$ is the inverse of the integration-by-substitution that recovers $\ln|f(x)|$ from $\dfrac{f'(x)}{f(x)}$ . Solving exponential equations is the algebraic counterpart of these analytic operations.
Section 10 — Numerical methods: transcendental equations like $e^x = 5x$ or $x e^x = 1$ have no closed-form solution. Candidates must locate roots via sign-change arguments and refine using Newton-Raphson or fixed-point iteration. The "first attempt analytically, then numerically" mindset is examined explicitly on Paper 1.
Mechanics (9MA0-03) — exponential decay with damping: Newton's law of cooling $\dfrac{d\theta}{dt} = -k(\theta - \theta_{env})$ integrates to $\theta - \theta_{env} = (\theta_0 - \theta_{env})e^{-kt}$ . Solving for $t$ given a target temperature is a direct application of the techniques in this lesson, dressed up as a Mechanics problem.

Mark-scheme literacy

Exponential-equation questions on 9MA0 split AO marks more evenly than surd questions because the modelling component is intrinsic:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Taking logs correctly, applying log laws, substituting $u = e^x$ , factorising the resulting quadratic, computing exact answers via $\ln$
AO2 (reasoning / interpretation)	20–30%	Rejecting $u < 0$ with explicit justification; choosing between $\ln$ and $\log_{10}$ ; interpreting "exact form" vs "to 3 s.f."; recognising hidden quadratics
AO3 (problem-solving / modelling)	15–25%	Translating word problems (half-life, doubling time, population decay) into exponential form; setting up the equation before solving

Examiner-rewarded phrasing: "since $e^x > 0$ for all real $x$ , we reject the negative root"; "taking natural logarithms of both sides, valid since both sides are positive"; "in exact form, $x = \dfrac{\ln 7}{\ln 3}$ "; "to 3 significant figures, $x \approx 1.77$ ". Phrases that lose marks: silently discarding negative roots from substitution quadratics; switching between $\log$ and $\ln$ mid-solution without comment; giving a decimal where exact form is demanded; computing $\ln(-2)$ without flagging it as undefined.

A specific Edexcel pattern to watch: questions phrased "find the exact value" demand a logarithm in the answer, never a decimal. If the natural answer comes out as $\ln 7 / \ln 3 \approx 1.7712$ , the exact form mark requires $\dfrac{\ln 7}{\ln 3}$ (or equivalently $\log_3 7$ ), and the decimal earns nothing additional. Conversely "to 3 s.f." demands the decimal — leaving $\dfrac{\ln 7}{\ln 3}$ unevaluated misses the accuracy mark.

Grade-band model answers

3-mark question

Question: Solve $3^{x+1} = 20$ , giving $x$ to 3 significant figures.

Grade C response (~210 words):

Take logs of both sides:

$\log(3^{x+1}) = \log 20$

Bring the exponent down: $(x + 1)\log 3 = \log 20$ .

So $x + 1 = \dfrac{\log 20}{\log 3} = \dfrac{1.301}{0.4771} = 2.727$ .

Hence $x = 2.727 - 1 = 1.727$ , which is $1.73$ to 3 s.f.

Examiner commentary: Full marks (3/3). The candidate correctly applies the power law of logarithms to bring $(x + 1)$ down from the exponent, computes the ratio of logs cleanly, and rounds to 3 s.f. as instructed. The choice of $\log_{10}$ is fine — $\ln$ would give the same numerical answer because the ratio $\log_b 20 / \log_b 3$ is independent of base $b$ (this is the change-of-base formula in disguise). One small presentation point: the candidate could have shown $\dfrac{\ln 20}{\ln 3}$ as an exact intermediate before evaluating, which would protect against arithmetic slips and signal log-law fluency. But the question asks for a decimal, so the numerical path is fully appropriate.

Grade A response (~340 words):*

Taking natural logarithms of both sides — valid because $3^{x+1} > 0$ and $20 > 0$ :

$\ln(3^{x+1}) = \ln 20$

Apply the power law $\ln(a^b) = b \ln a$ :

$(x + 1) \ln 3 = \ln 20$

Solve for $x + 1$ :

$x + 1 = \dfrac{\ln 20}{\ln 3}$

Solving Exponential Equations

Solving Exponential Equations

Type 1: Equations Reducible to a Common Base

Type 2: Taking Logarithms of Both Sides

Type 3: Hidden Quadratics

Type 4: Equations Involving eˣ

Type 5: Equations Mixing Exponentials and Constants

Checking Your Solutions

Summary

A-Level Deep Dive: Solving Exponential Equations

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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