Home / IGCSE Add Maths 0606 / Logarithmic and exponential functions / Solving exponential and logarithmic equations
Cambridge IGCSE · Topic 6.3
IGCSE Add Maths 0606 / Topic 6 / IGCSE

Bring the power down.

When the unknown sits in a power, as in \(3^{x} = 7\), take logarithms of both sides and the power comes down as a multiplier: \(x \lg 3 = \lg 7\). The same idea solves \(3^{2x + 1} = 5^{x - 2}\), \(500e^{0.2t} = 2000\) and, run backwards, equations such as \(\lg x + \lg(x - 3) = 1\).

Topic 6All logarithmic and exponential functions lessons →
The key idea

To solve \(a^{x} = b\), take \(\lg\) or \(\ln\) of both sides and use \(\log a^{x} = x\log a\): \(x = \dfrac{\lg b}{\lg a} = \dfrac{\ln b}{\ln a} = \log_{a} b\). On Paper 1 (no calculator) leave the answer exact; on Paper 2 give it to 3 significant figures. With \(e\), use \(\ln\): \(e^{kx} = c\) gives \(x = \dfrac{\ln c}{k}\). To solve an equation IN logs, combine the logs into a single log, convert \(\log_{a} X = n\) to \(X = a^{n}\), and reject any value that makes the argument of a log zero or negative.

x y O 1 1 2 7 1.77 (0, 1) y = 3x y = 7 3x = 7 take lg of both sides x lg 3 = lg 7 x = lg 7 ÷ lg 3 exact: x = log3 7 3 s.f.: x = 1.77
Fig. 1: the curve y = 3^x meets the line y = 7 where 3^x = 7. Taking logs gives x = lg 7 / lg 3 = log_3 7, which is 1.77 to 3 s.f. The curve passes through (0, 1) and is always above the x-axis, so 3^x = b has a solution only when b > 0
Section 01

Take logs of both sides.

You met \(\log_{a} b\) as 'the power of \(a\) that gives \(b\)'. When an equation has the unknown in the power, take \(\lg\) (or \(\ln\)) of both sides and use the power law \(\log a^{x} = x \log a\) to bring the power down. Either base works, as long as you use the SAME one on both sides: \(\dfrac{\lg 7}{\lg 3}\) and \(\dfrac{\ln 7}{\ln 3}\) are the same number, \(\log_{3} 7\). Fig. 1 shows the answer as the \(x\)-coordinate where \(y = 3^{x}\) meets \(y = 7\).

Worked example A single power
Solve \(3^{x} = 7\), giving (a) the exact answer, (b) the answer to 3 significant figures.
  1. Take \(\lg\) of both sides: \(\lg 3^{x} = \lg 7\).
  2. Power law: \(x \lg 3 = \lg 7\).
  3. (a) Divide: \(x = \dfrac{\lg 7}{\lg 3}\), which can also be written \(\dfrac{\ln 7}{\ln 3}\) or \(\log_{3} 7\).
  4. (b) \(x = \dfrac{0.8451\ldots}{0.4771\ldots} = 1.7712\ldots\). Check: \(3^{1.77} \approx 7\), and \(3^{1} < 7 < 3^{2}\), so an answer between 1 and 2 makes sense.
Answer: (a) \(x = \dfrac{\ln 7}{\ln 3} = \log_{3} 7\) (b) \(x = 1.77\)
Worked example An expression in the power
Solve \(5^{2x - 1} = 40\), giving your answer to 3 significant figures.
  1. Take \(\lg\): \((2x - 1)\lg 5 = \lg 40\). The bracket keeps the whole power together.
  2. Divide by \(\lg 5\): \(2x - 1 = \dfrac{\lg 40}{\lg 5} = 2.2920\ldots\)
  3. Add 1 and halve: \(2x = 3.2920\ldots\), so \(x = 1.6460\ldots\)
  4. Round only at the end: \(x = 1.65\).
Answer: \(x = 1.65\) (3 s.f.)
Your turn
Solve \(2^{x} = 15\), giving the exact answer and the answer to 3 significant figures.
Examiner trap

\(\dfrac{\lg 7}{\lg 3}\) is NOT \(\lg \dfrac{7}{3}\): the quotient law is about the log of a quotient, \(\lg 7 - \lg 3\), not a quotient of logs. \(\dfrac{\lg 7}{\lg 3} = 1.77\) but \(\lg \dfrac{7}{3} = 0.368\). Also, never 'cancel the lg' to get \(\dfrac{7}{3}\).

Section 02

Unknowns in both powers, and base e.

If \(x\) appears in two powers with different bases, take logs, expand the brackets, collect every \(x\) term on one side and factorise out \(x\). With base \(e\), take \(\ln\), because \(\ln e^{kx} = kx\). Growth and decay models such as \(N = 500e^{0.2t}\) are solved the same way: divide first so the exponential stands alone, then take \(\ln\). (If the equation has both \(e^{2x}\) and \(e^{x}\), it is a quadratic in \(e^{x}\): use the substitution from Section 4.3 instead.)

Worked example Two different bases
Solve \(3^{2x + 1} = 5^{x - 2}\), giving (a) an exact answer, (b) the answer to 3 significant figures.
  1. Take \(\ln\): \((2x + 1)\ln 3 = (x - 2)\ln 5\).
  2. Expand: \(2x\ln 3 + \ln 3 = x\ln 5 - 2\ln 5\).
  3. Collect the \(x\) terms: \(2x\ln 3 - x\ln 5 = -2\ln 5 - \ln 3\), so \(x(2\ln 3 - \ln 5) = -(2\ln 5 + \ln 3)\).
  4. Tidy with the log laws: \(2\ln 3 - \ln 5 = \ln \dfrac{9}{5}\) and \(2\ln 5 + \ln 3 = \ln 75\), so \(x = -\dfrac{\ln 75}{\ln 1.8}\).
  5. \(x = -\dfrac{4.3175\ldots}{0.5878\ldots} = -7.3453\ldots\)
Answer: (a) \(x = -\dfrac{\ln 75}{\ln 1.8}\) (b) \(x = -7.35\)
Worked example A growth model
The number of bacteria in a culture after \(t\) hours is \(N = 500e^{0.2t}\). Find the time when \(N = 2000\), giving (a) an exact answer, (b) the answer to 3 significant figures.
  1. \(500e^{0.2t} = 2000\). Divide by 500 first: \(e^{0.2t} = 4\).
  2. Take \(\ln\): \(0.2t = \ln 4\).
  3. (a) \(t = \dfrac{\ln 4}{0.2} = 5\ln 4\), which is also \(10\ln 2\).
  4. (b) \(t = 5 \times 1.3862\ldots = 6.931\ldots\) hours.
Answer: (a) \(t = 5\ln 4\) (b) \(t = 6.93\) hours
Your turn
Solve \(2e^{3x} = 9\), giving the exact answer and the answer to 3 significant figures.
Examiner trap

Do not take \(\ln\) before the exponential stands alone. \(\ln(500e^{0.2t})\) is \(\ln 500 + 0.2t\), NOT \(500 \times 0.2t\). Divide by 500 first and the step is clean. In \(3^{2x + 1} = 5^{x - 2}\), forgetting the brackets (writing \(2x + 1\ln 3\)) loses the multiplier on the \(2x\) and gives a wrong answer.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Which is the exact solution of \(2^{x} = 9\)?

Quick check+10 XP

Solve \(4^{x} = 30\), giving your answer to 3 significant figures.

Quick check+10 XP

Solve \(\lg x + \lg(x + 3) = 1\).

Quick check+10 XP

Solve \(e^{0.5t} = 6\) exactly.

Section 03

Equations in logarithms.

Now run the method backwards. When the unknown is inside the logs, use the laws of logarithms to combine everything into ONE log, then convert: \(\lg X = n\) means \(X = 10^{n}\), and \(\log_{a} X = n\) means \(X = a^{n}\). If the logs have different bases, change them to one base first, for example \(\log_{4} x = \dfrac{\log_{2} x}{\log_{2} 4} = \dfrac{1}{2}\log_{2} x\). Finally check every answer: the argument of each log in the ORIGINAL equation must be positive.

Worked example Combine, convert, reject
Solve \(\lg x + \lg(x - 3) = 1\).
  1. Product law: \(\lg[x(x - 3)] = 1\).
  2. Convert: \(x(x - 3) = 10^{1} = 10\), so \(x^{2} - 3x - 10 = 0\).
  3. Factorise: \((x - 5)(x + 2) = 0\), so \(x = 5\) or \(x = -2\).
  4. \(x = -2\) would need \(\lg(-2)\), which does not exist: reject it. Check \(x = 5\): \(\lg 5 + \lg 2 = \lg 10 = 1\).
Answer: \(x = 5\) only
Worked example Simultaneous equations with indices and logs
Solve the simultaneous equations \(2^{x} \times 4^{y} = 128\) and \(\lg x + \lg y = \lg 6\).
  1. Write everything as a power of 2: \(4^{y} = 2^{2y}\) and \(128 = 2^{7}\), so \(2^{x + 2y} = 2^{7}\) and \(x + 2y = 7\).
  2. Second equation: \(\lg xy = \lg 6\), so \(xy = 6\), with \(x > 0\) and \(y > 0\).
  3. Substitute \(x = 7 - 2y\): \((7 - 2y)y = 6\), so \(2y^{2} - 7y + 6 = 0\), and \((2y - 3)(y - 2) = 0\).
  4. \(y = 2\) gives \(x = 3\); \(y = \dfrac{3}{2}\) gives \(x = 4\). Both pairs are positive, so both are valid.
Answer: \(x = 3,\ y = 2\) or \(x = 4,\ y = \dfrac{3}{2}\)
Your turn
Solve \(\log_{2} x + \log_{4} x = 6\).
Examiner trap

\(\lg x + \lg(x - 3)\) is \(\lg[x(x - 3)]\), not \(\lg x \times \lg(x - 3)\) and not \(\lg(2x - 3)\). And always test your roots in the original equation: \(x = -2\) solves the quadratic but not the log equation. Do not reject a root just because it is small or a fraction; reject it only if a log argument is zero or negative.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to IGCSE Additional Mathematics Paper 1 and Paper 2 standard.

Finish the four checks above to unlock the exam questions
Exam style+20 XP

Solve \(2^{x + 1} = 3^{x}\), giving your answer to 3 significant figures.

Exam style+20 XP

Without a calculator, solve \(\log_{3} x + \log_{9} x = 3\).

Exam style+20 XP

Solve \(\lg(2x + 1) - \lg(x - 1) = 1\).

Exam style+20 XP

The mass, \(m\) grams, of a radioactive sample after \(t\) days is \(m = 80e^{-0.04t}\). Find the time for the mass to fall to 20 grams, to 3 significant figures.

Skill unlocked

Powers down, logs combined: the topic is complete.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics
Go deeper · practice
Six original Cambridge-style questions
\(6^{x} = 50\), unknowns in two powers, a log equation with two valid roots, \(\log_{3} x - 4\log_{x} 3 = 3\), simultaneous equations with indices and logs, and a population growth model.