Solving Equations using Logarithms
When an exponential equation cannot be solved by writing both sides with the same base, logarithms find the answer. Taking logs of \(3^x=20\) gives \(x\log 3=\log 20\), so \(x\approx2.73\). In NSW Year 10 Advanced maths you solve these equations with a calculator and use them in growth, decay and finance problems (NESA outcome MA5‑LOG‑P‑01).
You will solve equations with brackets, negative and decimal indices in the power, rearranging first so the power stands alone. You will find when a growing population or a fading dose reaches a set level, how many whole years an investment takes to pass a target, and when a vehicle's value drops below a set amount. Harder questions put the unknown in powers of two different bases, or hide a quadratic in the powers. It is part of Year 10 Advanced Logarithms and Polynomials.
Every question with a fully worked solution.
- Solving Equations using Logarithms - Video - Logarithms : Equations Watch
Theory
To solve an exponential equation such as \(3^x = 20\), make the power the subject, take logs of both sides and bring the index down: \(x = \dfrac{\log 20}{\log 3}\). This Year 10 Advanced maths page covers brackets in the index, growth and decay models, compound interest to a whole number of years, unknowns on both sides and hidden quadratics.
An exponential equation has the unknown in the index, such as \(2^x = 10\). Unless both sides are powers of the same number, guess-and-check is slow; logarithms solve it in one step.
Taking logs of both sides keeps the equation balanced. The power law then brings the index down: \(\log 2^x = x\log 2\). Any base works, but base \(10\) is on every calculator, so \(x = \dfrac{\log_{10} 10}{\log_{10} 2} \approx 3.32\).
Equivalently, \(2^x = 10\) means \(x = \log_2 10\), and the change of base rule gives the same division.
Models such as \(A = P(1 + r)^n\) and \(N = N_0 \times b^{t}\) lead to these equations. When the answer must be a whole number of years, solve an inequality and check the two whole numbers either side.
This comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
| Model | Formula | Multiplier each period |
|---|---|---|
| Compound interest | \(A = P(1 + r)^n\) | \(1 + r\), e.g. \(1.035\) |
| Depreciation | \(S = V_0(1 - r)^n\) | \(1 - r\), e.g. \(0.88\) |
| Doubling every \(d\) hours | \(N = N_0 \times 2^{\genfrac{}{}{}{1}{t}{d}}\) | \(2\) per \(d\) hours |
| Halving every \(h\) hours | \(A = A_0 \times 0.5^{\genfrac{}{}{}{1}{t}{h}}\) | \(0.5\) per \(h\) hours |
How to solve an exponential equation
- Isolate the power: divide or subtract so it stands alone. \(7 \times 2^x = 91\) becomes \(2^x = 13\).
- Take logs of both sides and bring the whole index down, in brackets if it has more than one term: \(\log 4^{x + 1} = (x + 1)\log 4\).
- Solve the linear equation that is left. Treat \(\log 4\) as an ordinary number.
- Round only at the end, and check by substituting back.
Other forms
- Negative or decimal index: \(2^{-x} = 0.05\) gives \(-x\log 2 = \log 0.05\). \(5^{0.02x} = 3\) gives \(0.02x\log 5 = \log 3\); divide by \(0.02\) last.
- Exact answers: if both sides are powers of the same number, compare indices instead: \(2^{-x} = \dfrac{1}{8}\) gives \(x = 3\).
- Different bases both sides: take logs, expand the brackets, collect the \(x\) terms and factorise out \(x\) (Example 4).
- Comparing two investments: divide one side by the other to get a single power, \(\left(\dfrac{1.07}{1.05}\right)^n\), then take logs.
- Whole periods: solve the inequality, round up to the next whole number, and check the values either side (Example 3).
(i) \(10^x = 4.7\) (four decimal places)
(ii) \(5 \times 3^{2x - 1} = 80\) (two decimal places)
(i) Write in log form straight away.
| \(x\) | \(=\) | \(\log_{10} 4.7\) |
| \(\) | \(=\) | \(0.67209\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{0.6721}\) |
(ii) Divide by \(5\) first so the power stands alone. Then take logs and keep the bracket.
| \(3^{2x - 1}\) | \(=\) | \(16\) |
| \((2x - 1)\log_{10} 3\) | \(=\) | \(\log_{10} 16\) |
| \(2x - 1\) | \(=\) | \(\dfrac{\log_{10} 16}{\log_{10} 3} = 2.5237\ldots\) |
| \(2x\) | \(=\) | \(3.5237\ldots\) |
| \(x\) | \(=\) | \(1.7618\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1.76}\) |
Do not multiply \(5 \times 3\) first: the index belongs only to the \(3\).
Substitute \(P = 3000\) and isolate the power.
| \(1200 \times 1.08^t\) | \(=\) | \(3000\) |
| \(1.08^t\) | \(=\) | \(2.5\) |
| \(t\log_{10} 1.08\) | \(=\) | \(\log_{10} 2.5\) |
| \(t\) | \(=\) | \(\dfrac{\log_{10} 2.5}{\log_{10} 1.08}\) |
| \(\) | \(=\) | \(11.9059\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{11.9}\text{ years}\) |
The population reaches \(3000\) about \(11.9\) years after 2020, during 2031.
(i) \(\$6000\) is invested at \(5.2\%\) p.a., compounded annually. After how many whole years is it first worth more than \(\$10\,000\)?
(ii) A machine bought for \(\$30\,000\) loses \(20\%\) of its value each year. After how many whole years is it first worth less than \(\$8000\)?
(i) Use \(A = P(1 + r)^n\) as an inequality.
| \(6000(1.052)^n\) | \(\gt\) | \(10\,000\) |
| \(1.052^n\) | \(\gt\) | \(\dfrac{5}{3}\) |
| \(n\) | \(\gt\) | \(\dfrac{\log_{10} \genfrac{}{}{}{1}{5}{3}}{\log_{10} 1.052} = 10.07\ldots\) |
Check: after \(10\) years it is worth \(\$9961.13\), after \(11\) years \(\$10\,479.11\). Answer: \(\mathbf{11}\) years.
(ii) Each year the value is multiplied by \(1 - 0.2 = 0.8\).
| \(30\,000(0.8)^n\) | \(\lt\) | \(8000\) |
| \(0.8^n\) | \(\lt\) | \(\dfrac{4}{15}\) |
| \(n\log_{10} 0.8\) | \(\lt\) | \(\log_{10} \dfrac{4}{15}\) |
| \(n\) | \(\gt\) | \(\dfrac{\log_{10} \genfrac{}{}{}{1}{4}{15}}{\log_{10} 0.8} = 5.92\ldots\) |
The sign reverses because \(\log_{10} 0.8\) is negative and you divided by it.
Check: after \(5\) years \(\$9830.40\), after \(6\) years \(\$7864.32\). Answer: \(\mathbf{6}\) years.
(i) \(5^x = 3^{x + 2}\)
(ii) \(9^x - 4 \times 3^x - 5 = 0\)
(i) Take logs, expand, and collect the \(x\) terms on one side.
| \(x\log_{10} 5\) | \(=\) | \((x + 2)\log_{10} 3\) |
| \(x\log_{10} 5 - x\log_{10} 3\) | \(=\) | \(2\log_{10} 3\) |
| \(x(\log_{10} 5 - \log_{10} 3)\) | \(=\) | \(2\log_{10} 3\) |
| \(x\) | \(=\) | \(\dfrac{2\log_{10} 3}{\log_{10} 5 - \log_{10} 3}\) |
| \(\) | \(=\) | \(4.3013\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{4.30}\) |
(ii) \(9^x = (3^2)^x = (3^x)^2\), so let \(u = 3^x\).
| \(u^2 - 4u - 5\) | \(=\) | \(0\) |
| \((u - 5)(u + 1)\) | \(=\) | \(0\) |
| \(u\) | \(=\) | \(5 \text{ or } -1\) |
\(3^x\) is always positive, so \(3^x = -1\) has no solution.
| \(3^x\) | \(=\) | \(5\) |
| \(x\) | \(=\) | \(\dfrac{\log_{10} 5}{\log_{10} 3} = 1.4649\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1.46}\) |
Common pitfalls
Frequently asked questions
How do you solve an equation with x in the exponent?
Make the power the subject, take the log of both sides, bring the index down in front using the power law, then divide. For \(2^x = 10\), \(x = \dfrac{\log 10}{\log 2} \approx 3.32\).
Does it matter which log base I use?
No, as long as you use the same base on both sides. Base 10 is easiest because it is on every calculator.
How do you find how many years for an investment to reach a target?
Write the compound interest formula as an inequality, isolate the power, take logs and solve for n. Round up to the next whole year and check the values either side.
Why does the inequality sign change when solving with logs?
The log of a number between 0 and 1 is negative. Dividing both sides of an inequality by a negative number reverses the sign.
How do you solve 4 to the x minus 2 to the x equals 6?
Notice \(4^x = (2^x)^2\). Let \(u = 2^x\), solve \(u^2 - u - 6 = 0\) to get \(u = 3\) or \(u = -2\), reject the negative, then solve \(2^x = 3\) with logs.