Solve an exponential equation

(unknown in the power, or in the base)

In 2 of 20 recent exams
Paper 1 or 2
2 to 3 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator), 2–3 marks, inside a functions question. Rare as a part of its own (2 of 20 exam versions: one Paper 1 (no calculator), one Paper 2 (calculator)), but it is the solving step inside many other questions: a model's rate constant or time, a stationary point of an exponential function, each root of a quadratic in $e^x$ (Where exponents and logs hide). On Paper 2 (calculator) the exact form is still found by hand.

Spot it in the exam

Exam phrases to spot (4)

“find the exact value of $x$ ”
“give your answer in the form $a + \ln b$ ”
“show that the value of $a$ is …”
“solve $f(x) = g(x)$ ”

In disguise (4)

$f(x) = g(x)$ where $g$ is a transformation of $f$ : the same exponential is on both sides
two exponentials and no constant term: they collect into one (a common factor that is never $0$ )
a point with a fractional $x$ -coordinate on the graph of $y = a^x$ : the unknown is the base
the zero, or $x$ -intercept, of one exponential and a constant, such as $y = 4 - e^{2x}$ : $f(x) = 0$ is an equation to solve here (with no constant term, the intercept is Exponential and log graphs)

Variants (2)

collect the exponential terms, then take $\ln$ : the exact answer $a + \ln b$ (, by hand)
unknown base with a fractional exponent: $a^{\frac{m}{n}} = c$ , so raise both sides to $\frac{n}{m}$ (, "show that")

Looks like it, but isn't (5)

two powers of $e^x$ and a constant, such as $e^{2x} - 5e^x + 6 = 0$ : Quadratic in disguise
a percentage per period with no formula in $t$ , and a whole number of periods ("least $n$ "): Find n: which term, how many terms, least n (a money question reaches it through the time row of Compound interest and depreciation, which sends the time unknown to Find n)
a model in a context, such as a rate constant from a half-life, or the time to reach a value, even when a formula in continuous time asks for the first whole unit: Exponential and log models
logs in the equation: Solve a logarithmic equation (the same numbers, the opposite first move:
the $y$ -intercept, asymptote or range of an exponential graph, and an $x$ -intercept of an exponential expression with no constant term (factor out the lowest power of $e^x$ ): Exponential and log graphs. The same equation, with no graph in the question, stays here.

Worked example

The base from a point, then a common factor

Paper 1 (no calculator)

Question

Let $f(x) = a^x$ , where $a > 0$ . The graph of $f$ passes through the point $\left(-\frac{3}{2}, \frac{1}{27}\right)$ .
(a) Show that $a = 9$ . [2]
(b) Solve $f(x) = 7 \times 3^x$ , giving your answer in the form $\dfrac{\ln p}{\ln q}$ , where $p, q \in \mathbb{Z}^+$ . [3]

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Which method?

The equation	Method
Unknown in the exponent: one exponential after collecting, or one on each side with different bases ( $2^{x+3} = 5^x$ )	take $\ln$ (or $\log$ ) of both sides; with two bases, bracket each exponent and collect the $x$ terms (or divide first to get one power: $8 \times 2^x = 5^x$ gives $\left(\frac{5}{2}\right)^x = 8$ )
Both sides are powers of one prime	common base, then equate the exponents (as in Exact value of a logarithm)
Several exponentials, no constant term	factor out, or divide by, the common exponential (it is never $0$ ), then take $\ln$
Two powers of $e^x$ and a constant	Quadratic in disguise
Unknown in the base: $x^{\frac{m}{n}} = c$	raise both sides to the power $\frac{n}{m}$

The same numbers, the unknown in different places: $x^{\frac{3}{2}} = 8$ has the unknown in the base, so $x = 8^{\frac{2}{3}} = 4$ . $\left(\frac{3}{2}\right)^x = 8$ has it in the power, so $x = \frac{\ln 8}{\ln 1.5} = 5.128\ldots \approx 5.13$ .

Steps

Collect
the exponential terms. First write exponentials whose bases are powers of the same prime in one base: $9^x = 3^{2x} = \left(3^x\right)^2$ . With no constant term, factor out the common exponential or divide through by it: it is never $0$ . With one exponential on each side and different bases, leave them there and go to step 3.
Isolate one exponential equal to a constant
(or one on each side, as in step 1). If the constant is $0$ or negative, there is no solution ( $e^x > 0$ for every $x$ ).
Take $\ln$
(or $\log$ ) of both sides and solve the linear equation. With two different bases, bracket each exponent when the power law brings it down: $\ln 2^{x+3} = (x + 3)\ln 2$ .
Give the exact form
asked for, $a + \ln b$ or a quotient of logs. On the decimal comes after the exact form.
Unknown in the base:
isolate the power, $x^{\frac{m}{n}} = c$ with $c > 0$ (the fraction in lowest terms), and raise both sides to the reciprocal power $\frac{n}{m}$ . If $m$ is odd there is one root. If $m$ is even, the negative number works too, because an even power hides the sign: $x^{\frac{2}{3}} = 9$ gives $x = \pm 27$ , but $x^{\frac{3}{2}} = 27$ gives only $x = 9$ . Keep the root the conditions allow (a base $a > 0$ keeps $27$ ).
Check
by substituting back; on , with a graph (zeros and intersections).

Common mistakes

Taking $\ln$ of a sum term by term
$\ln\left(2e^{x - 3} + 10\right)$ is not $\ln 2 + (x - 3) + \ln 10$ . Collect the exponentials into one term first.
Cancelling an exponential without a reason
dividing by $3^x$ or $e^x$ is allowed because it is never $0$ ; say so. (Dividing by $x$ is different: $x$ can be $0$ , and dividing loses that root. Factorise instead, unless $x = 0$ is already ruled out, as it is when the question says $x > 0$ .)
Solving $e^x = -2$
there is no solution, because $e^x > 0$ : say so. " $x = \ln(-2)$ " is not a solution; left in the answer, it loses the for rejecting it (Quadratic in disguise).

More mistakes (6)

Merging a number into a power
$7 \times 3^x$ is not $21^x$ , so $\ln\left(7 \times 3^x\right) = \ln 7 + x\ln 3$ .
A decimal where an exact form is asked for,
even on : $4.61$ scores nothing for " $a + \ln b$ ".
An exact form left unsimplified, or in the wrong form
$x = \frac{\ln 6}{0.5}$ must be written $2\ln 6$ ; a markscheme refused the unsimplified quotient as a final answer. And $\log_3 7$ is exact, but "the form $\frac{\ln p}{\ln q}$ " needs $\frac{\ln 7}{\ln 3}$ .
The wrong reciprocal power for an unknown base
$a^{\frac{3}{2}} = 27$ gives $a = 27^{\frac{2}{3}} = 9$ , not $27^{\frac{3}{2}}$ .
An exponent left unbracketed
$\ln 2^{x+3} = (x + 3)\ln 2$ , not $x + 3\ln 2$ .
Checking instead of showing
in "show that $a = 9$ ", substituting $9$ and getting $\frac{1}{27}$ is a check, not a proof. Start from the point, $a^{-\frac{3}{2}} = \frac{1}{27}$ , and end on $a = 9$ (the LHS-to-RHS layout).

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