Unlimited generated questions on this topic, at up to four difficulty levels, with full mark schemes. Notes and worksheets are open to everyone. Practice needs a free account.
Or take a quick quiz: 10 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
Solving equations where the unknown is in the power, such as \(a^x = b\), by taking logarithms of both sides. Harder questions have different bases on each side or are quadratics in disguise, such as \(9^x - 4(3^x) - 21 = 0\). These are common 3 to 6 mark questions.
Key facts
\(a^x = b \Rightarrow x = \log_a b = \dfrac{\ln b}{\ln a}\)
\(\ln(a^x) = x\ln a\)
Change of base: \(\log_a b = \dfrac{\log_c b}{\log_c a}\)
\(a^{x+k} = a^k \times a^x\) and \(a^{x-k} = \dfrac{a^x}{a^k}\)
\(a^x > 0\) always, so a root \(u \leq 0\) of the quadratic gives no solution
Notes
The basic method
To solve \(a^x = b\) (where \(b > 0\)), take logs of both sides: \(x\log a = \log b\), so \(x = \dfrac{\log b}{\log a}\)
You can use ln or \(\log_{10}\), as long as you use the same on both sides. Use ln whenever \(\mathrm{e}\) is involved.
The answer can also be written \(x = \log_a b\). For example, \(5^x = 40\) gives \(x = \log_5 40 = \dfrac{\ln 40}{\ln 5} = 2.29\) (3 s.f.).
\(a^x = b\) has no solution if \(b \leq 0\), because \(a^x\) is always positive.
Harder linear equations
For \(a^{px + q} = b\), the whole bracket comes down: \((px + q)\ln a = \ln b\)
With different bases, e.g. \(3^{2x} = 7^{x + 1}\):\(2x\ln 3 = (x + 1)\ln 7\). Expand, collect the \(x\) terms and factorise: \(x(2\ln 3 - \ln 7) = \ln 7\), so \(x = \dfrac{\ln 7}{2\ln 3 - \ln 7}\)
Keep \(\ln 3\), \(\ln 7\) and so on exact until the final line, to avoid rounding errors.
Disguised quadratics
diagram
Write each term using one exponential: \(4^x = (2^x)^2\),\(2^{x+1} = 2 \times 2^x\),\(3^{x-1} = \dfrac{3^x}{3}\), \(\mathrm{e}^{2x} = (\mathrm{e}^x)^2\)
Substitute \(u = 2^x\) (or \(u = \mathrm{e}^x\)), solve the quadratic in \(u\), then solve \(2^x = u\) for each root.
Reject any root with \(u \leq 0\), because \(2^x > 0\) for all \(x\). Write this reason in your answer.
\(3^x = 7\) has one solution, \(x = \dfrac{\ln 7}{\ln 3} = 1.77\); \(3^x = -3\) has none, because \(3^x > 0\)
If there is an \(\mathrm{e}^{-x}\) term, multiply every term by \(\mathrm{e}^x\) first (\(\mathrm{e}^x \times \mathrm{e}^{-x} = 1\)).
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own, dM = depends on the earlier method mark.
Solve \(a^{px+q}=b\) to a given accuracy
2 to 3 marksGrade D
Take logs of both sides and bring the power down as a bracket.
Rearrange for \(x\), keeping the logs exact until the end.
Round to the accuracy asked for (often 3 significant figures).
Example. Solve \(3^{2x - 1} = 20\), giving your answer to 3 significant figures.
Take logs of both sides, using brackets for powers with more than one term.
Expand the brackets and collect all the \(x\) terms on one side.
Factorise out \(x\) and divide. Simplify using the log laws if an exact form is asked for.
Example. Solve \(2^{x + 3} = 5^x\), giving your answer (a) in the form \(x = \dfrac{\ln a}{\ln b}\), where \(a\) and \(b\) are rational numbers, (b) to 3 significant figures.
Show the model answerHide the model answer
(a) \((x + 3)\ln 2 = x\ln 5\)
M1
\(3\ln 2 = x(\ln 5 - \ln 2)\)
M1
\(x = \dfrac{\ln 8}{\ln 2.5}\)
A1
(b) \(x = 2.27\)
A1
Solve a disguised quadratic
4 to 6 marksGrade A
Rewrite every term in terms of one exponential, e.g. \(9^x = (3^x)^2\)
Let \(u = 3^x\) and solve the quadratic in \(u\)
Reject any value of \(u\) that is zero or negative, giving the reason.
Solve \(3^x = u\) for the remaining value(s).
Example. Solve \(9^x - 4(3^x) - 21 = 0\), giving your answer to 3 significant figures.
Show the model answerHide the model answer
\(9^x = (3^x)^2\), so with \(u = 3^x\): \(u^2 - 4u - 21 = 0\)
B1
\((u - 7)(u + 3) = 0\), so \(u = 7\) or \(u = -3\)
M1 A1
\(3^x = -3\) has no solution because \(3^x > 0\); \(3^x = 7 \Rightarrow x = \dfrac{\ln 7}{\ln 3}\)
M1
\(x = 1.77\)
A1
Solve an equation in \(\mathrm{e}^x\) and \(\mathrm{e}^{-x}\)
4 marksGrade A
Multiply every term by \(\mathrm{e}^x\), remembering \(\mathrm{e}^x \times \mathrm{e}^{-x} = 1\)
Rearrange into a quadratic in \(\mathrm{e}^x\) and factorise.
Reject a negative value of \(\mathrm{e}^x\), then take ln.
Example. Solve \(\mathrm{e}^x - 6\mathrm{e}^{-x} = 1\), giving your answer in an exact form.
Solve \(a^{px+q}=b\)Bring the whole bracket down: \((px + q)\ln a = \ln b\), then rearrange.
Grade B
Solve equations with different basesE.g. \(2^{x+3} = 5^x\): take logs, expand, collect the \(x\) terms and factorise.
Grade A
Solve disguised quadratics in \(a^x\)Write \(9^x = (3^x)^2\), substitute \(u = 3^x\), solve, then reject any \(u \leq 0\)
Grade A
Solve equations with \(\mathrm{e}^x\) and \(\mathrm{e}^{-x}\)Multiply every term by \(\mathrm{e}^x\) to get a quadratic in \(\mathrm{e}^x\)
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
(a) Solve \(7^{3x} = 50\), giving your answer to 3 significant figures.[2]
(b) Find the exact solution of \(\mathrm{e}^{3x + 4} = 5\).[2]
(c) Find the exact solution of \(\ln(3x + 3) = 1\).[2]
Show the answer and mark scheme
(a)Answer: \(x = 0.670\)
M1 for taking logarithms: \(3x\ln 7 = \ln 50\) or \(3x = \log_{7} 50\)
A1 for awrt \(0.670\)
Worked solution: \(3x = \log_{7} 50 = \frac{\ln 50}{\ln 7} = 2.0103\ldots\), so \(x = 0.670\).
(b)Answer: \(x = \frac{\ln 5 - 4}{3}\)
M1 for \(3x + 4 = \ln 5\)
A1 for \(x = \frac{\ln 5 - 4}{3}\) oe
Worked solution: \(3x + 4 = \ln 5 \Rightarrow x = \frac{\ln 5 - 4}{3}\) (\(\approx -0.797\)).
(c)Answer: \(x = \frac{\mathrm{e} - 3}{3}\)
M1 for \(3x + 3 = \mathrm{e}\)
A1 for \(x = \frac{\mathrm{e} - 3}{3}\) oe
Worked solution: \(3x + 3 = \mathrm{e} \Rightarrow x = \frac{\mathrm{e} - 3}{3}\) (\(\approx -0.094\)).
Question 2Medium5 marks
(a) Show that the solution of the equation \[3^{x + 1} = 5^{2x}\] can be written in the form \(x = \dfrac{\ln P}{\ln Q}\), where \(P\) and \(Q\) are rational numbers to be found.[4]
(b) Hence find the value of \(x\), giving your answer to 3 significant figures.[1]