Exponential Equations and Inequalities

When the unknown is in the index, as in $5^x = 12$ or $3^{x+1} = 4^{2x-1}$, no amount of ordinary algebra will bring it down. Taking logarithms will: the power law $\ln(a^x) = x\ln a$ turns the index into a coefficient, and the equation becomes linear. The syllabus asks you to use logarithms to solve equations and inequalities in which the unknown appears in indices, quoting examples such as $2^x < 5$, $3 \times 2^{3x-1} < 5$ and $3^{x+1} = 4^{2x-1}$. These appear on almost every P3 paper, sometimes alone for three or four marks, sometimes hidden inside a quadratic, a modulus or a growth model.

Why taking logs works

The power law of logarithms says $\log(a^k) = k\log a$, for any base of logarithm. Applied to $a^x = b$:

$$a^x = b \quad\Rightarrow\quad \ln(a^x) = \ln b \quad\Rightarrow\quad x\ln a = \ln b \quad\Rightarrow\quad x = \frac{\ln b}{\ln a}.$$

Two facts make this legitimate:

• Both sides are positive (any $a^x$ with $a > 0$ is positive), so their logarithms exist.
• $\ln$ is one-one: if $p = q$ then $\ln p = \ln q$, and conversely. Taking logs of an equation neither gains nor loses solutions.

Any base works. $\lg$ (base $10$) and $\ln$ give the same final answer, because $\dfrac{\lg b}{\lg a} = \dfrac{\ln b}{\ln a}$. Use $\ln$ by default, since it is what the rest of P3 uses.

Equations with the unknown in both indices

For $a^{px + q} = b^{rx + s}$, take logs and bring both powers down. The result is a linear equation in $x$, with the logs acting as numerical coefficients.

For example, $8^{x} = 4^{x+1}$ is $2^{3x} = 2^{2x + 2}$, so $3x = 2x + 2$ and $x = 2$.

Hidden quadratics

If an equation contains both $a^{2x}$ and $a^x$, it is a quadratic in disguise, because $a^{2x} = \left(a^x\right)^2$. Substitute $u = a^x$. Watch for the index laws that produce these terms:

$$a^{2x} = (a^x)^2, \qquad a^{x+1} = a\cdot a^x, \qquad a^{x-1} = \frac{a^x}{a}, \qquad a^{2x+1} = a\left(a^x\right)^2, \qquad 4^x = \left(2^x\right)^2.$$

Solve for $u$, then solve $a^x = u$ for each positive root. A negative or zero value of $u$ gives no solution, because $a^x > 0$. This is the same technique as equations that are quadratic in a function of x from P1.

Inequalities

The syllabus examples $2^x < 5$ and $3 \times 2^{3x-1} < 5$ are inequalities. The method is the same as for equations, with one point of care.

Taking $\ln$ of both sides is safe, because $\ln$ is an increasing function: if $0 < p < q$ then $\ln p < \ln q$. The direction of the inequality is kept.

The danger is the next step, when you divide by $\ln a$. If $a > 1$, then $\ln a > 0$ and the sign is kept. If $0 < a < 1$, then $\ln a < 0$, and dividing by a negative number reverses the inequality.

$$2^x < 5 \Rightarrow x\ln 2 < \ln 5 \Rightarrow x < \frac{\ln 5}{\ln 2} \qquad(\ln 2 > 0)$$ $$0.8^x < 0.1 \Rightarrow x\ln 0.8 < \ln 0.1 \Rightarrow x > \frac{\ln 0.1}{\ln 0.8} \qquad(\ln 0.8 < 0,\ \text{so the sign turns})$$

The second result makes sense: $0.8^x$ decreases as $x$ increases, so it falls below $0.1$ only for large enough $x$ (here $x > 10.3$).

The smallest integer that works

A common context: "Find the smallest integer $n$ such that $1.04^n > 3$." Solve as an inequality, $n > \dfrac{\ln 3}{\ln 1.04} = 28.01\ldots$, then choose the next integer: $n = 29$. Do not round $28.01$ to $28$; check that $1.04^{28} = 2.9987 < 3$ while $1.04^{29} = 3.1187 > 3$.

Equations with e

Equations in $e$ follow the same rules, but $\ln e^{f(x)} = f(x)$ directly, so the step "divide by $\ln a$" disappears: $e^{2x - 1} = 5 \Rightarrow 2x - 1 = \ln 5$. See Euler's number, e^x and ln x for these and for quadratics in $e^x$.

Worked examples

Practice

Answers

1. $x = \dfrac{\ln 20}{\ln 7} = 1.54$ (3 s.f.).

2. $(3x - 1)\ln 2 = \ln 5 \Rightarrow 3x - 1 = \dfrac{\ln 5}{\ln 2} = 2.3219\ldots \Rightarrow x = 1.11$ (3 s.f.).

3. $2x\ln 5 = (x + 2)\ln 3 \Rightarrow x(2\ln 5 - \ln 3) = 2\ln 3 \Rightarrow x = \dfrac{2\ln 3}{2\ln 5 - \ln 3} = \dfrac{\ln 9}{\ln\frac{25}{3}} = 1.04$ (3 s.f.).

4. $x\ln 2 < \ln 5$ and $\ln 2 > 0$, so $x < \dfrac{\ln 5}{\ln 2}$, i.e. $x < 2.32$ (3 s.f.).

5. $x\ln 0.6 < \ln 0.05$. Since $\ln 0.6 < 0$, the inequality reverses: $x > \dfrac{\ln 0.05}{\ln 0.6} = 5.86$ (3 s.f.).

6. $n\ln 1.04 > \ln 3 \Rightarrow n > \dfrac{\ln 3}{\ln 1.04} = 28.01\ldots$, so $n = 29$. (Check: $1.04^{28} = 2.9987 < 3$ and $1.04^{29} = 3.1187 > 3$.)

7. $4^x = \left(2^x\right)^2$ and $3\left(2^{x+1}\right) = 6\left(2^x\right)$. With $u = 2^x$: $u^2 - 6u - 16 = 0 \Rightarrow (u - 8)(u + 2) = 0$. $u = -2$ is impossible since $2^x > 0$, so $2^x = 8$ and $x = 3$.

8. With $u = 2^x$: $u^2 - 5u + 4 < 0 \Rightarrow (u - 1)(u - 4) < 0 \Rightarrow 1 < u < 4$. So $2^0 < 2^x < 2^2$, giving $0 < x < 2$.

9. (a) Let $u = 2^x > 0$, so $3\left(2^{-x}\right) = \dfrac{3}{u}$ and the equation is $u|4 - u| = 3$. If $u \le 4$: $u(4 - u) = 3 \Rightarrow u^2 - 4u + 3 = 0 \Rightarrow (u - 1)(u - 3) = 0$, so $u = 1$ or $u = 3$, both $\le 4$. These give $x = 0$ and $x = \dfrac{\ln 3}{\ln 2}$. If $u > 4$: $u(u - 4) = 3 \Rightarrow u^2 - 4u - 3 = 0 \Rightarrow u = 2 \pm \sqrt{7}$. Only $u = 2 + \sqrt{7} \approx 4.65$ satisfies $u > 4$, giving $x = \dfrac{\ln(2 + \sqrt{7})}{\ln 2}$ ($\approx 2.22$). Three solutions: $x = 0$, $x = \dfrac{\ln 3}{\ln 2}$ ($\approx 1.58$) and $x = \dfrac{\ln(2 + \sqrt{7})}{\ln 2}$. (b) With $u = 3^x$: $u^2 - 2u - 15 > 0 \Rightarrow (u - 5)(u + 3) > 0 \Rightarrow u > 5$ or $u < -3$. Since $u = 3^x > 0$, only $3^x > 5$ is possible, so $x > \dfrac{\ln 5}{\ln 3} = 1.46$ (3 s.f.).