9709 P3 §3.2 — Solving Exponential Equations

💡 Lead In

Log Laws Review Quiz

Before we solve exponential equations, quickly check you remember the log laws from Lesson 1 — especially the Power Law, because it lets us bring an exponent "down" in front of the log, and it is exactly what the change-of-base formula depends on. Answer each question to check.

1. Simplify $\log_3(3^5)$. Not answered

$5$

$3$

$15$

$243$

Power Law: $\log_c(m^n) = n\log_c m$, so $\log_3(3^5) = 5\cdot\log_3 3 = 5 \cdot 1 = 5$.

2. Evaluate $\log 2 + \log 5$. Not answered

$\log 7$

$10$

$1$

$7$

Product Law: $\log_c(mn) = \log_c m + \log_c n$, so $\log 2 + \log 5 = \log(2\cdot 5) = \log 10 = 1$.

3. Evaluate $\log 100 - \log 10$. Not answered

$90$

$1$

$2$

$10$

Quotient Law: $\log_c\left(\dfrac{m}{n}\right) = \log_c m - \log_c n$, so $\log 100 - \log 10 = \log\left(\dfrac{100}{10}\right) = \log 10 = 1$.

4. To solve $a^x = b$ for $x$, which of these expresses $x$? Not answered

$x = \log_a b$

$x = \log_b a$

$x = a^b$

$x = ab$

By definition, $a^x = b \iff x = \log_a b$. This is how taking logs brings the variable out of the exponent.

Exponential & Logarithmic Graphs Review

Recall the shapes of the exponential and logarithmic functions from earlier work, and their domains and ranges. The three graphs below will be used in the questions:

(i)

(ii)

(iii)

1. Graph (i) shows which function? Not answered

$y = 2^x$

$y = \log_2 x$

$y = 2^{-x}$

$y = \dfrac{1}{x}$

Graph (i) rises through the point $(0,1)$ and grows as $x$ increases — this is the increasing exponential $y = 2^x$.

2. For $y = 2^x$, what are its domain and range? Not answered

Domain $\mathbb{R}$, Range $y > 0$

Domain $x > 0$, Range $\mathbb{R}$

Domain $\mathbb{R}$, Range $y \ge 1$

Domain $\mathbb{R}$, Range $\mathbb{R}$

Every real $x$ is allowed, and $2^x > 0$ for all $x$ (the graph approaches but never touches $y = 0$). So domain $\mathbb{R}$, range $y > 0$.

3. For $y = \log_2 x$, what are its domain and range? Not answered

Domain $\mathbb{R}$, Range $y > 0$

Domain $x \ge 1$, Range $\mathbb{R}$

Domain $x > 0$, Range $\mathbb{R}$

Domain $\mathbb{R}$, Range $\mathbb{R}$

You can only take the log of a positive number, so $x > 0$; the output $y$ can be any real number. Domain $x > 0$, range $\mathbb{R}$.

4. How are the graphs of $y = 2^x$ and $y = \log_2 x$ related? Not answered

They are identical graphs

Mirror images in the line $y = x$

Mirror images in the $x$-axis

Parallel lines with the same slope

Since the log is the inverse of the exponential ($a^x = b \iff x = \log_a b$), their graphs are reflections of each other in the line $y = x$.

Sketch It — exponential and logarithmic functions

Sketch each curve with your mouse or finger (click and drag to draw). Then press Check to compare your sketch with the standard graph. Try to capture the key features: the asymptote, the intercepts, and whether the curve is increasing or decreasing.

Draw $y=e^x$, then check.

Draw $y=\ln x$, then check.

Draw $y=0.5^x$, then check.

Draw $y=\log_{0.5} x$, then check.

The Wall

Now try this one: solve $3^{x+1} = 4^{2x-1}$. The bases are different — you cannot write 3 as a nice power of 4 (or vice versa) the way you could with $2^x = 8 = 2^3$. Guessing will not work, because the answer is not an integer. The two exponential curves below cross at a single point, but where exactly?

The Need

When the bases are different, the only way to bring the variable out of the exponent is to take logarithms of both sides. Today we extend your log-law toolkit to solve exactly this kind of equation — and the inequalities that come with it.

📖 Key Concepts

Key relation 1 — Taking logs of both sides:

You may apply a logarithm of the same base to both sides of an equation — for example $\log$ (base 10) or $\ln$ (base $e$) — and the two sides remain equal:

$a^x = b \;\Longleftrightarrow\; \log_c(a^x) = \log_c b \;\Longleftrightarrow\; x\log_c a = \log_c b$

Combined with the Power Law $\log_c(m^n) = n\log_c m$, taking logs brings the variable $x$ down from the exponent, turning an exponential equation into a linear one. Any base works, as long as the same base is used on both sides.

Key relation 2 — Change of base (derivation):

When a calculator has no direct button for the base $a$, we convert to a base it can handle. Let $y = \log_a b$, so $a^y = b$. Take logs to any base $c$ of both sides and apply the Power Law:

$y\log_c a = \log_c b \;\Longrightarrow\; y = \dfrac{\log_c b}{\log_c a}$

Change of base: $\log_a b = \dfrac{\log_c b}{\log_c a}$ (for any base $c$ with $c > 0$, $c \neq 1$)

Choosing $c = 10$ or $c = e$ gives the forms a calculator can evaluate: $\log_a b = \dfrac{\log b}{\log a} = \dfrac{\ln b}{\ln a}$.

Key relation 3 — Substitution method:

For an equation of the form $a^{2x} + p\,a^x + q = 0$, use substitution to reduce it to a quadratic. Follow the steps in order:

Substitute: let $u = a^x$ (with $u > 0$, since $a^x > 0$). Then $a^{2x} = (a^x)^2 = u^2$.
Rewrite: the equation becomes the quadratic $\;u^2 + p\,u + q = 0$.
Solve for $u$ by factorising or the quadratic formula; keep only the root(s) with $u > 0$.
Back-substitute: for each valid $u$, solve $a^x = u$ for $x = \log_a u$.

Activity 1 · Group Equation Relay Board

Group relay (10 min): Form groups of 4. Each member handles one step of solving $3^{x+1} = 4^{2x-1}$. The board unlocks step by step — each member must get their step correct before the next member's step appears. Work together to reach the final answer!

Team Score: 0 / 40

Step 1 · Member A — "Take Logs" Answer to unlock Step 2

Solve $3^{x+1} = 4^{2x-1}$. What is the correct equation after taking ln of both sides?

$(x+1)\ln 3 = (2x-1)\ln 4$

$(x+1)\log 3 = (2x-1)\ln 4$

$x \cdot \ln 3 = 2x \cdot \ln 4$

$\ln 3^{(x+1)} + \ln 4^{(2x-1)} = 0$

Step 2 · Member B — "Expand" Locked — complete Step 1 first

Expand both sides of $(x+1)\ln 3 = (2x-1)\ln 4$ using the distributive law.

$x \cdot \ln 3 + \ln 3 = 2x \cdot \ln 4 - \ln 4$

$x \cdot \ln 3 + \ln 3 = 2x \cdot \ln 4 + \ln 4$

$x \cdot \ln 3 - \ln 3 = 2x \cdot \ln 4 - \ln 4$

$(x+1) \cdot \ln 3 = (2x-1) \cdot \ln 4$

Step 3 · Member C — "Collect" Locked — complete Step 2 first

Rearrange to collect all x-terms on the left side.

$x \cdot \ln 3 - 2x \cdot \ln 4 = -\ln 4 - \ln 3$

$x \cdot \ln 3 + 2x \cdot \ln 4 = -\ln 4 + \ln 3$

$x(\ln 3 - \ln 4) = -(\ln 4 + \ln 3)$

$x \cdot \ln 3 = 2x \cdot \ln 4 - \ln 4 - \ln 3$

Step 4 · Member D — "Solve" Locked — complete Step 3 first

Factor out x and solve for x.

$x = \dfrac{\ln 4 + \ln 3}{2 \ln 4 - \ln 3}$

$x = \dfrac{\ln 4 - \ln 3}{2 \ln 4 + \ln 3}$

$x = \dfrac{-(\ln 4 + \ln 3)}{2 \ln 4 - \ln 3}$

$x = \dfrac{\ln 3 + \ln 4}{\ln 4 - 2\ln 3}$

The key to solving $3^{x+1} = 4^{2x-1}$ is taking logs first, then carefully expanding, collecting x-terms, and factoring. Each step builds on the last — one slip and the whole chain breaks!

Worked Example 1 · Group Solving $3^{(2x+1)} = 20$ by Taking Logs

Exercise 2B Q9(vi) Standard

Use logarithms to the base 10 to solve the following equation: $3^{(2x+1)} = 20$

Step 1 — Take $\log$ of both sides

$\log(3^{2x+1}) = \log(20)$

Step 2 — Apply the power law

$(2x+1)\log 3 = \log 20$

Step 3 — Isolate $2x+1$

$2x+1 = \dfrac{\log 20}{\log 3}$

Step 4 — Solve for $x$

$2x+1 \approx \dfrac{1.3010}{0.4771} \approx 2.7268$

$2x \approx 1.7268$

$x \approx 0.8634$

Answer

$x \approx 0.863$ (3 s.f.)

Worked Example 2 · Individual Substitution Method: $2^{2x} + 3(2^x) - 18 = 0$

Exercise 2B Q10(i) Standard

Solve the equation: $2^{2x} + 3(2^x) - 18 = 0$

Step 1 — Substitute $u = 2^x$

Since $2^{2x} = (2^x)^2 = u^2$, the equation becomes:

$u^2 + 3u - 18 = 0$

Step 2 — Factor the quadratic

$u^2 + 3u - 18 = (u + 6)(u - 3) = 0$

Step 3 — Solve for $u$

$u = -6$ or $u = 3$

Since $u = 2^x > 0$, reject $u = -6$. So $u = 3$.

Step 4 — Back-substitute

$2^x = 3$

$x = \log_2 3 = \dfrac{\log 3}{\log 2} \approx \dfrac{0.4771}{0.3010} \approx 1.585$

Answer

$x \approx 1.585$ (3 s.f.)

Worked Example 3 · Group Substitution with a Leading Term: $3^{(2x+1)} + 5(3^x) - 2 = 0$

Exercise 2B Q10(v) Standard

Solve the equation: $3^{(2x+1)} + 5(3^x) - 2 = 0$

Step 1 — Rewrite $3^{2x+1}$

$3^{2x+1} = 3 \cdot 3^{2x} = 3(3^x)^2$

Step 2 — Substitute $u = 3^x$

The equation becomes $3u^2 + 5u - 2 = 0$

Step 3 — Factor the quadratic

$3u^2 + 5u - 2 = (3u - 1)(u + 2) = 0$

Step 4 — Solve for $u$

$u = \dfrac{1}{3}$ or $u = -2$

Since $u = 3^x > 0$, reject $u = -2$. So $u = \dfrac{1}{3}$.

Step 5 — Back-substitute

$3^x = \dfrac{1}{3} = 3^{-1}$

$x = -1$

Answer

$x = -1$

Worked Example 4 · Group Solving $0.6^x < 0.8$

Exercise 2B Q7(iv) Standard

Solve the inequality $0.6^x < 0.8$, giving your answer correct to 3 significant figures.

Step 1 — Take $\log$ of both sides

$\log(0.6^x) < \log 0.8$

Step 2 — Apply the power law

$x\log 0.6 < \log 0.8$

Step 3 — Evaluate each $\log$ before dividing

$\log 0.6 \approx -0.2218$ (negative) and $\log 0.8 \approx -0.0969$

Since $\log 0.6 < 0$, dividing by it reverses the inequality:

$x > \dfrac{\log 0.8}{\log 0.6}$

Step 4 — Evaluate

$x > \dfrac{-0.0969}{-0.2218} \approx 0.437$

Answer

$x > 0.437$ (3 s.f.)

Worked Example 5 · Individual Solving $|2^x - 4| < 2$

Exercise 2B Q7(ix) Standard

Solve the inequality $|2^x - 4| < 2$, giving your answer correct to 3 significant figures.

Step 1 — Remove the modulus

$|2^x - 4| < 2 \implies -2 < 2^x - 4 < 2$

Step 2 — Isolate $2^x$

Add 4 throughout: $2 < 2^x < 6$

Step 3 — Take logs

Since $\log$ is increasing, the inequality direction is preserved:

$1 < x < \log_2 6$

Step 4 — Evaluate with change of base

$\log_2 6 = \dfrac{\log 6}{\log 2} \approx \dfrac{0.7782}{0.3010} \approx 2.585$

Answer

$1 < x < 2.59$ (3 s.f.)

Worked Example 6 · Group Solving $2|3^x - 1| = 3^x$

Exercise 2B Q16 Advanced

Solve the equation $2|3^x - 1| = 3^x$, giving your answers correct to 3 significant figures.

Step 1 — Consider the two cases for the modulus

Case 1: $3^x - 1 \ge 0$ (i.e. $x \ge 0$)

Case 2: $3^x - 1 < 0$ (i.e. $x < 0$)

Step 2 — Case 1: $x \ge 0$

$2(3^x - 1) = 3^x \implies 2\cdot 3^x - 2 = 3^x \implies 3^x = 2$

$x = \log_3 2 = \dfrac{\log 2}{\log 3} \approx \dfrac{0.3010}{0.4771} \approx 0.631$ ✓ (since $x \ge 0$)

Step 3 — Case 2: $x < 0$

$2(1 - 3^x) = 3^x \implies 2 - 2\cdot 3^x = 3^x \implies 3^x = \dfrac{2}{3}$

$x = \log_3\!\left(\dfrac{2}{3}\right) = \dfrac{\log(2/3)}{\log 3} \approx \dfrac{-0.1761}{0.4771} \approx -0.369$ ✓ (since $x < 0$)

Answer

$x \approx -0.369$ or $x \approx 0.631$ (3 s.f.)

Activity 2 · Individual Error Hunter: Exponential Edition

Individual challenge (8 min): Three worked solutions contain exactly one error each. Click on the step that contains the error. Identify all three errors to become a certified Error Hunter!

Errors found: 0 / 3 | Score: 0

Problem 1. Solve $2^x = 5$ Click the step with the error

Step 1 Take log: $x \cdot \log 2 = \log 5$

Step 2 Rearrange: $x = \dfrac{\log 2}{\log 5}$

Step 3 Calculate: $x = \dfrac{0.301}{0.699} \approx 0.431$

Problem 2. Solve $3 \times 2^{3x-1} = 5$ Click the step with the error

Step 1 Divide by 3: $2^{3x-1} = \dfrac{5}{3}$

Step 2 Take log: $(3x-1) \cdot \log 2 = \log\left(\dfrac{5}{3}\right)$

Step 3 Expand: $3x \cdot \log 2 - 1 = \log\left(\dfrac{5}{3}\right)$

Step 4 Solve: $3x \cdot \log 2 = 1 + \log\left(\dfrac{5}{3}\right)$, so $x = \dfrac{1 + \log\left(\dfrac{5}{3}\right)}{3 \log 2} \approx 1.353$

Problem 3. Solve $e^{2x} - 8e^x + 16 = 0$ Click the step with the error

Step 1 Substitute $u = e^x$: $u^2 - 8u + 16 = 0$

Step 2 Factor: $(u - 8)(u - 2) = 0$

Step 3 Solutions: $u = 8$ or $u = 2$, so $e^x = 8$ or $e^x = 2$

Step 4 Final answers: $x = \ln 8$ or $x = \ln 2$

Common traps: (1) inverting the log fraction, (2) forgetting that $\log$ multiplies every term inside a bracket, (3) careless factorisation. Always check each algebraic step by substituting back or re-expanding.

🎯 Concept Quiz

Each question targets one lesson objective. Answer all 4 to complete the quiz.

Score: 0 / 4

1. Solve $2^x = 7$, giving $x$ correct to 3 significant figures. Not answered

$x \approx 1.94$

$x \approx 2.65$

$x \approx 2.81$

$x \approx 3.00$

Take $\log$ of both sides: $x \log 2 = \log 7$, so $x = \dfrac{\log 7}{\log 2} \approx \dfrac{0.8451}{0.3010} \approx 2.81$.

2. Which expression is equivalent to $\log_5 12$ using base-10 logarithms? Not answered

$\log 12 - \log 5$

$\dfrac{\log 12}{\log 5}$

$\dfrac{\log 5}{\log 12}$

$\log 12 \times \log 5$

Change of base formula: $\log_a b = \dfrac{\log_c b}{\log_c a}$ for any positive base $c \neq 1$. With $c = 10$: $\log_5 12 = \dfrac{\log 12}{\log 5}$.

3. Using $u = 3^x$ to solve $3^{2x} - 4(3^x) + 3 = 0$, what is the resulting quadratic equation in $u$? Not answered

$u^2 + 4u + 3 = 0$

$u^2 - 4u + 3 = 0$

$u^2 - 4u - 3 = 0$

$2u^2 - 4u + 3 = 0$

Since $3^{2x} = (3^x)^2 = u^2$ and $3^x = u$, the equation $3^{2x} - 4(3^x) + 3 = 0$ becomes $u^2 - 4u + 3 = 0$.

4. Solve the inequality $2^x < 5$. Not answered

$x < 2.5$

$x < \dfrac{5}{2}$

$x < \log_2 5 \approx 2.322$

$x > \log_2 5$

Take $\log_2$ of both sides (log is increasing, so the inequality direction is preserved): $x < \log_2 5 = \dfrac{\log 5}{\log 2} \approx \dfrac{0.6990}{0.3010} \approx 2.322$.

🎉

You scored 0 / 4

📝 More Practice

Solve each problem on paper, then click Hint or Reveal Solution to check your work.

Exercise 2B Q9(vii) Standard

Use logarithms to the base 10 to solve: $2^{(5x-2)} = 30$

Hint

Take $\log$ of both sides and isolate $5x - 2$ before solving for $x$.

Step 1 — Take $\log$ of both sides

$(5x-2)\log 2 = \log 30$

Step 2 — Isolate $5x-2$

$5x - 2 = \dfrac{\log 30}{\log 2} \approx \dfrac{1.4771}{0.3010} \approx 4.9069$

Step 3 — Solve for $x$

$5x \approx 6.9069$

$x \approx 1.381$

Answer

$x \approx 1.381$ (3 s.f.)

Exercise 2B Q9(viii) Medium

Use logarithms to the base 10 to solve: $2^{3x} = 5^{(2x+1)}$

Hint

Take $\log$ of both sides, apply the power law to bring $3x$ and $(2x+1)$ down as multipliers, then collect $x$-terms on one side.

Step 1 — Take $\log$ of both sides

$3x \log 2 = (2x+1)\log 5$

Step 2 — Expand and collect $x$-terms

$3x \log 2 = 2x \log 5 + \log 5$

$x(3\log 2 - 2\log 5) = \log 5$

Step 3 — Solve for $x$

$x = \dfrac{\log 5}{3\log 2 - 2\log 5}$

$x \approx \dfrac{0.6990}{3(0.3010) - 2(0.6990)} = \dfrac{0.6990}{0.9030 - 1.3980} = \dfrac{0.6990}{-0.4950}$

$x \approx -1.412$

Answer

$x \approx -1.412$ (3 s.f.)

Exercise 2B Q10(ii) Standard

Solve the equation: $5^{2x} + 5^x - 2 = 0$

Hint

Let $u = 5^x$ (with $u > 0$). Then $5^{2x} = u^2$, giving a quadratic in $u$. Solve and back-substitute.

Step 1 — Substitute $u = 5^x$

$u^2 + u - 2 = 0$

Step 2 — Factor the quadratic

$u^2 + u - 2 = (u + 2)(u - 1) = 0$

Step 3 — Solve for $u$

$u = -2$ (rejected since $u = 5^x > 0$) or $u = 1$.

Step 4 — Back-substitute

$5^x = 1 = 5^0$

$x = 0$

Check

$5^{2(0)} + 5^0 - 2 = 1 + 1 - 2 = 0$ ✓

Answer

$x = 0$

Exercise 2B Q10(vi) Medium

Solve the equation: $2(7^{2x}) - 7^{(x+1)} + 3 = 0$

Hint

Rewrite as $2(7^x)^2 - 7(7^x) + 3 = 0$ and substitute $u = 7^x$ (with $u > 0$).

Step 1 — Substitute $u = 7^x$

$2u^2 - 7u + 3 = 0$

Step 2 — Factor the quadratic

$2u^2 - 7u + 3 = (2u - 1)(u - 3) = 0$

Step 3 — Solve for $u$

$u = \dfrac{1}{2}$ or $u = 3$

Step 4 — Back-substitute

$7^x = \dfrac{1}{2} \implies x = \log_7\!\left(\dfrac{1}{2}\right)$

$7^x = 3 \implies x = \log_7 3$

Step 5 — Evaluate with change of base

$x = \dfrac{\log(1/2)}{\log 7} \approx \dfrac{-0.3010}{0.8451} \approx -0.356$

$x = \dfrac{\log 3}{\log 7} \approx \dfrac{0.4771}{0.8451} \approx 0.565$

Answer

$x \approx -0.356$ or $x \approx 0.565$ (3 s.f.)

Exercise 2B Q7(x) Standard

Solve the inequality $|5^x - 7| < 4$, giving your answer correct to 3 significant figures.

Hint

Remove the modulus to get $-4 < 5^x - 7 < 4$, then isolate $5^x$ and take $\log$.

Step 1 — Remove the modulus

$|5^x - 7| < 4 \implies -4 < 5^x - 7 < 4$

Step 2 — Isolate $5^x$

Add 7 throughout: $3 < 5^x < 11$

Step 3 — Take $\log$ of all three sides

$\log 3 < x\log 5 < \log 11$

Step 4 — Divide by $\log 5$ (positive)

$\dfrac{\log 3}{\log 5} < x < \dfrac{\log 11}{\log 5}$

Step 5 — Evaluate

$\dfrac{0.4771}{0.6990} < x < \dfrac{1.0414}{0.6990}$

$0.683 < x < 1.49$

Answer

$0.683 < x < 1.49$ (3 s.f.)

Exercise 2B Q17 Advanced

Using the substitution $u = 3^x$, or otherwise, solve, correct to 3 significant figures, the equation $3^x = 2 + 3^{-x}$

Hint

With $u = 3^x$, note that $3^{-x} = \dfrac{1}{u}$. Multiply through by $u$ to form a quadratic.

Step 1 — Substitute $u = 3^x$

$u = 2 + \dfrac{1}{u}$

Step 2 — Multiply through by $u$ (since $u > 0$)

$u^2 = 2u + 1 \implies u^2 - 2u - 1 = 0$

Step 3 — Solve the quadratic

$u = \dfrac{2 \pm \sqrt{4 + 4}}{2} = \dfrac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$

Since $u = 3^x > 0$, reject $1 - \sqrt{2}$ and keep $u = 1 + \sqrt{2}$.

Step 4 — Back-substitute

$3^x = 1 + \sqrt{2}$

$x = \log_3(1 + \sqrt{2}) = \dfrac{\log(1 + \sqrt{2})}{\log 3}$

Step 5 — Evaluate

$x \approx \dfrac{\log 2.4142}{0.4771} \approx \dfrac{0.3828}{0.4771} \approx 0.802$

Answer

$x \approx 0.802$ (3 s.f.)

🎫 Exit Ticket

Before you leave, tick the objectives you can confidently do on your own.

I can solve $a^x = b$ by taking logarithms of both sides (e.g. $3^{2x+1} = 20$). I can use the change of base formula $\log_a b = \dfrac{\log b}{\log a}$ to evaluate logarithms with arbitrary bases. I can solve equations of the form $a^{2x} + p\,a^x + q = 0$ using the substitution $u = a^x$. I can solve logarithmic equations by combining log terms and checking the validity of solutions.