Exponentials and logarithms
Exponentials and logarithms
- A-level
- Calculator allowed
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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The curve $y = 2^x$ is drawn. State the coordinates of the point where it meets the $y$-axis, and state the equation of the horizontal asymptote. Write down the range of the function.
3 marks
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Simplify $\log_2 32 - \log_2 4 + \log_2 \dfrac{1}{2}$.
3 marks
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Write $e^{\ln 7}$ as an integer, and write $\ln (e^4)$ as an integer. Explain the relationship you use.
4 marks
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Solve $\log_2 x + \log_2 (x - 2) = 3$.
4 marks
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Solve $e^{2x} - 4e^x + 3 = 0$.
5 marks
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The curve $y = e^{2x} - 4$ is drawn. State the equation of its horizontal asymptote, and find the coordinates of the points where it meets the axes.
5 marks
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A population is modelled by $P = 200e^{0.04t}$, where $t$ is the time in years after the start of the model. Find the exact population when $t = 10$, and the exact time at which the population reaches $400$.
6 marks
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The number of bacteria in a culture is modelled by $N = N_0 e^{kt}$. The number doubles every $5$ hours. Show that $k = \dfrac{\ln 2}{5}$, and find the exact time taken for the culture to triple.
7 marks
Mark scheme
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Answer. The curve meets the $y$-axis at $(0, 1)$. The asymptote is $y = 0$. The range is $y > 0$.
- $2^0 = 1$, so the $y$-intercept is $(0, 1)$.
- $2^x$ approaches $0$ as $x \to -\infty$, but never reaches it.
- An exponential function with positive base is always positive.
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Answer. $2$
- $\log_2 32 = 5$ and $\log_2 4 = 2$.
- $\log_2 \dfrac{1}{2} = -1$.
- $5 - 2 - 1 = 2$.
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Answer. $e^{\ln 7} = 7$ and $\ln (e^4) = 4$. The exponential and logarithm undo each other.
- $\ln$ is the inverse of the function $e^x$.
- $e^{\ln 7} = 7$.
- $\ln (e^4) = 4$.
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Answer. $x = 4$
- $\log_2 \big(x(x - 2)\big) = 3$.
- $x(x - 2) = 8$, so $x^2 - 2x - 8 = 0$.
- $(x - 4)(x + 2) = 0$.
- $x = -2$ is not in the domain, so $x = 4$.
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Answer. $x = 0$ or $x = \ln 3$
- Let $u = e^x$. Then $u^2 - 4u + 3 = 0$.
- $(u - 1)(u - 3) = 0$, so $u = 1$ or $u = 3$.
- $e^x = 1$ gives $x = 0$.
- $e^x = 3$ gives $x = \ln 3$.
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Answer. The asymptote is $y = -4$. The curve meets the axes at $(0, -3)$ and $(\ln 2, 0)$.
- As $x \to -\infty$, $e^{2x} \to 0$, so the asymptote is $y = -4$.
- When $x = 0$, $y = 1 - 4 = -3$.
- When $y = 0$, $e^{2x} = 4$, so $2x = \ln 4 = 2\ln 2$ and $x = \ln 2$.
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Answer. When $t = 10$, $P = 200e^{0.4}$. The population reaches $400$ when $t = 25\ln 2$.
- Substitute $t = 10$ to get $200e^{0.4}$.
- $400 = 200e^{0.04t}$ gives $e^{0.04t} = 2$.
- $0.04t = \ln 2$.
- $t = \dfrac{\ln 2}{0.04} = 25\ln 2$.
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Answer. $k = \dfrac{\ln 2}{5}$, and the culture triples after $\dfrac{5\ln 3}{\ln 2}$ hours.
- After $5$ hours, $N = 2N_0$, so $e^{5k} = 2$.
- $5k = \ln 2$, so $k = \dfrac{\ln 2}{5}$.
- For a triple, $e^{kt} = 3$.
- $t = \dfrac{\ln 3}{k} = \dfrac{5\ln 3}{\ln 2}$.