Year 10 Maths - 10A Solving Exponential Equations Using Logarithms

Solving Exponential Equations Using Logarithms

Q: How do you solve 4 to the x minus 2 to the x equals 6?

Notice 4 to the x is 2 to the x squared. Let u equal 2 to the x, solve the quadratic u squared minus u minus 6 equals 0 to get u = 3 or u = negative 2, reject the negative, then solve 2 to the x equals 3 with logs.

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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 10A 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 prepares you for Mathematical Methods and Specialist Mathematics in Years 11 and 12.

An exponential equation asks where a curve reaches a height. Logs find that $x$ exactly.

Any base of log works in step 2, as long as you use the same one on both sides. Base $10$ is on every calculator.

\[a^x = b \;\Rightarrow\; x = \log_a b = \dfrac{\log_{10} b}{\log_{10} a}\]

a^{x} = b \Rightarrow x = \frac{\log_{10} b}{\log_{10} a}

\[\log a^{\,\text{(anything)}} = (\text{anything}) \times \log a\]

\log a^{k} = k \log a

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

Inequalities. Dividing by a log of a number less than \(1\), such as \(\log 0.8\), divides by a negative, so the inequality sign reverses.

Hidden quadratic. \(4^x = (2^x)^2\) and \(9^x = (3^x)^2\). Let \(u = 2^x\) or \(u = 3^x\), solve the quadratic, and reject any negative \(u\) because a power of a positive base is always positive.

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).

Example 1 — Taking logs of both sides

Solve, giving answers to the accuracy stated:
(i) $10^x = 4.7$ (four decimal places)
(ii) $5 \times 3^{2x - 1} = 80$ (two decimal places)

Solution

(i) Write in log form straight away.

$x$	$=$	$\log_{10} 4.7$
	$=$	$0.67209\ldots$
	$\approx$	$\mathbf{0.6721}$

x = \log_{10} 4.7 \approx 0.6721

(ii) Divide by $5$ first so the power stands alone. Then take logs and keep the bracket.

$3^{2x - 1}$	$=$	$16$
$(2x - 1)\log_{10} 3$	$=$	$\log_{10} 16$
$2x - 1$	$=$	$\dfrac{\log_{10} 16}{\log_{10} 3} = 2.5237\ldots$
$2x$	$=$	$3.5237\ldots$
$x$	$=$	$1.7618\ldots$
	$\approx$	$\mathbf{1.76}$

(2 x - 1) \log_{10} 3 = \log_{10} 16

Do not multiply $5 \times 3$ first: the index belongs only to the $3$.

Example 2 — A growth model

A town's population $t$ years after 2020 is modelled by $P = 1200 \times 1.08^t$. When will the population reach $3000$? Give $t$ correct to one decimal place.

Solution

Substitute $P = 3000$ and isolate the power.

$1200 \times 1.08^t$	$=$	$3000$
$1.08^t$	$=$	$2.5$
$t\log_{10} 1.08$	$=$	$\log_{10} 2.5$
$t$	$=$	$\dfrac{\log_{10} 2.5}{\log_{10} 1.08}$
	$=$	$11.9059\ldots$
	$\approx$	$\mathbf{11.9}\text{ years}$

t = \frac{\log_{10} 2.5}{\log_{10} 1.08} \approx 11.9

The population reaches $3000$ about $11.9$ years after 2020, during 2031.

Example 3 — Whole numbers of years

(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 $\$8000$?

Solution

(i) Use $A = P(1 + r)^n$ as an inequality.

$6000(1.052)^n$	$\gt$	$10\,000$
$1.052^n$	$\gt$	$\dfrac{5}{3}$
$n$	$\gt$	$\dfrac{\log_{10} \genfrac{}{}{}{1}{5}{3}}{\log_{10} 1.052} = 10.07\ldots$

n > \frac{\log_{10} \frac{5}{3}}{\log_{10} 1.052}

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 $1 - 0.2 = 0.8$.

$30\,000(0.8)^n$	$\lt$	$8000$
$0.8^n$	$\lt$	$\dfrac{4}{15}$
$n\log_{10} 0.8$	$\lt$	$\log_{10} \dfrac{4}{15}$
$n$	$\gt$	$\dfrac{\log_{10} \genfrac{}{}{}{1}{4}{15}}{\log_{10} 0.8} = 5.92\ldots$

n > \frac{\log_{10} \frac{4}{15}}{\log_{10} 0.8}

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.

Example 4 — Unknowns on both sides

Solve, correct to two decimal places:
(i) $5^x = 3^{x + 2}$
(ii) $9^x - 4 \times 3^x - 5 = 0$

Solution

(i) Take logs, expand, and collect the $x$ terms on one side.

$x\log_{10} 5$	$=$	$(x + 2)\log_{10} 3$
$x\log_{10} 5 - x\log_{10} 3$	$=$	$2\log_{10} 3$
$x(\log_{10} 5 - \log_{10} 3)$	$=$	$2\log_{10} 3$
$x$	$=$	$\dfrac{2\log_{10} 3}{\log_{10} 5 - \log_{10} 3}$
	$=$	$4.3013\ldots$
	$\approx$	$\mathbf{4.30}$

x = \frac{2 \log_{10} 3}{\log_{10} 5 - \log_{10} 3} \approx 4.30

(ii) $9^x = (3^2)^x = (3^x)^2$, so let $u = 3^x$.

$u^2 - 4u - 5$	$=$	$0$
$(u - 5)(u + 1)$	$=$	$0$
$u$	$=$	$5 \text{ or } -1$

u^{2} - 4 u - 5 = 0

$3^x$ is always positive, so $3^x = -1$ has no solution.

$3^x$	$=$	$5$
$x$	$=$	$\dfrac{\log_{10} 5}{\log_{10} 3} = 1.4649\ldots$
	$\approx$	$\mathbf{1.46}$

3^{x} = 5 \Rightarrow x \approx 1.46

Common pitfalls

Isolate before you log. $\log(5 \times 3^x)$ is $\log 5 + x\log 3$, not $x\log 15$. Dividing by $5$ first avoids the trap.

Bracket the index. $\log 3^{2x - 1} = (2x - 1)\log 3$. Without the bracket only the $-1$ gets multiplied.

Watch the sign flip. When you divide an inequality by the log of a number between $0$ and $1$, you are dividing by a negative.

Round up for 'first exceeds'. If $n \gt 9.2$, the answer is $10$ whole years, even though $9.2$ rounds to $9$.

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.

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More in Solving Exponential Equations Using Logarithms

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

\(x\)	\(=\)	\(\log_{10} 4.7\)
\(\)	\(=\)	\(0.67209\ldots\)
\(\)	\(\approx\)	\(\mathbf{0.6721}\)

\(3^{2x - 1}\)	\(=\)	\(16\)
\((2x - 1)\log_{10} 3\)	\(=\)	\(\log_{10} 16\)
\(2x - 1\)	\(=\)	\(\dfrac{\log_{10} 16}{\log_{10} 3} = 2.5237\ldots\)
\(2x\)	\(=\)	\(3.5237\ldots\)
\(x\)	\(=\)	\(1.7618\ldots\)
\(\)	\(\approx\)	\(\mathbf{1.76}\)

\(1200 \times 1.08^t\)	\(=\)	\(3000\)
\(1.08^t\)	\(=\)	\(2.5\)
\(t\log_{10} 1.08\)	\(=\)	\(\log_{10} 2.5\)
\(t\)	\(=\)	\(\dfrac{\log_{10} 2.5}{\log_{10} 1.08}\)
\(\)	\(=\)	\(11.9059\ldots\)
\(\)	\(\approx\)	\(\mathbf{11.9}\text{ years}\)

\(6000(1.052)^n\)	\(\gt\)	\(10\,000\)
\(1.052^n\)	\(\gt\)	\(\dfrac{5}{3}\)
\(n\)	\(\gt\)	\(\dfrac{\log_{10} \genfrac{}{}{}{1}{5}{3}}{\log_{10} 1.052} = 10.07\ldots\)

\(30\,000(0.8)^n\)	\(\lt\)	\(8000\)
\(0.8^n\)	\(\lt\)	\(\dfrac{4}{15}\)
\(n\log_{10} 0.8\)	\(\lt\)	\(\log_{10} \dfrac{4}{15}\)
\(n\)	\(\gt\)	\(\dfrac{\log_{10} \genfrac{}{}{}{1}{4}{15}}{\log_{10} 0.8} = 5.92\ldots\)

\(x\log_{10} 5\)	\(=\)	\((x + 2)\log_{10} 3\)
\(x\log_{10} 5 - x\log_{10} 3\)	\(=\)	\(2\log_{10} 3\)
\(x(\log_{10} 5 - \log_{10} 3)\)	\(=\)	\(2\log_{10} 3\)
\(x\)	\(=\)	\(\dfrac{2\log_{10} 3}{\log_{10} 5 - \log_{10} 3}\)
\(\)	\(=\)	\(4.3013\ldots\)
\(\)	\(\approx\)	\(\mathbf{4.30}\)

\(u^2 - 4u - 5\)	\(=\)	\(0\)
\((u - 5)(u + 1)\)	\(=\)	\(0\)
\(u\)	\(=\)	\(5 \text{ or } -1\)

\(3^x\)	\(=\)	\(5\)
\(x\)	\(=\)	\(\dfrac{\log_{10} 5}{\log_{10} 3} = 1.4649\ldots\)
\(\)	\(\approx\)	\(\mathbf{1.46}\)

Solving Exponential Equations Using Logarithms

📖 Theory

How to solve an exponential equation

Other forms

Common pitfalls

Frequently asked questions

Theory