e^x and ln x: graphs and equations

The graph of y = e^{ax + b} + c

y = e^x passes through (0, 1), is always positive and increasing, and has asymptote y = 0.

For y = e^{ax + b} + c: the asymptote is y = c and the y-intercept is e^b + c. If a < 0 the curve falls instead of rising. Do not stop the curve at the y-axis.

The graph of y = ln x

y = ln x is the reflection of y = e^x in y = x. It passes through (1, 0), has asymptote x = 0 and exists only for x > 0. y = ln(x - 2) needs x > 2.

Solving equations

• e^{ax + b} = p: take ln of both sides, ax + b = ln p.
• ln(ax + b) = q: take e of both sides, ax + b = e^q.
• e^2x - 5e^x + 6 = 0 is a quadratic in e^x. Reject e^x ≤ 0, since e^x is always positive.

Worked example

Solve e^{2x - 1} = 7. Give x to 3 significant figures.

1. 2x - 1 = ln 7
2. x = (1 + ln 7)/2

Answer: x = 1.47

Check yourself

1. Solve e^{3x + 1} = 20. Give x to 3 decimal places.
   Show answer

   0.665. 3x + 1 = ln 20, so x = (ln 20 - 1)/3 = 0.665.

2. Solve ln(2x - 3) = 1.5. Give x to 3 decimal places.
   Show answer

   3.741. 2x - 3 = e¹.5, so x = (e¹.5 + 3)/2 = 3.741.

3. What is the horizontal asymptote of y = 4 - 3e^-x?
   • y = 0
   • y = -3
   • y = 1
   • y = 4
   Show answer

   y = 4. As x gets large, e^-x gets close to 0, so y gets close to 4.

4. Find the y-coordinate where y = 5 - 2e^3x crosses the y-axis.
   Show answer

   3. At x = 0, e⁰ = 1, so y = 5 - 2 = 3.

5. Solve e^2x - 5e^x + 6 = 0.
   • x = 2 or x = 3
   • x = e² or x = e³
   • x = ln 2 or x = ln 3
   • x = -ln 2 or x = -ln 3
   Show answer

   x = ln 2 or x = ln 3. Let u = e^x: u² - 5u + 6 = 0, so u = 2 or 3. Then x = ln 2 or ln 3.

6. What is the largest possible domain of f(x) = ln(x - 2)?
   • x > 2
   • x ≥ 2
   • x > 0
   • x > -2
   Show answer

   x > 2. ln needs a positive input: x - 2 > 0.

7. Solve ln x + ln(x + 3) = ln 10.
   Show answer

   2. ln(x(x + 3)) = ln 10, so x² + 3x - 10 = 0 and (x + 5)(x - 2) = 0. x = -5 is rejected because ln(-5) is undefined.

8. f(x) = e^x + 3, x ∈ ℝ. Find f^-1(x).
   • ln(x + 3), x > -3
   • ln x - 3, x > 0
   • e^{x - 3}, x ∈ ℝ
   • ln(x - 3), x > 3
   Show answer

   ln(x - 3), x > 3. y = e^x + 3 gives e^x = y - 3, so x = ln(y - 3). The range of f is f(x) > 3.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Pure Mathematics 3 scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.