A-Level Exponential functions Practice Questions

Free A-Level Exponential functions practice questions with full step-by-step worked solutions. Covers powers, evaluating exponentials, negative indices, exponentials. Practise exam-style problems and check your method.

powersevaluating exponentialsnegative indicesexponentialspowers of fractionsexponential constant
A-Level70 questionsStep-by-step solutions
Question 1
1 markeasy
Evaluate 343^{4}.
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Worked solution

  1. Write out the power as repeated multiplication

    34=3×3×3×33^{4}=3\times 3\times 3\times 3

    A power tells us how many times to multiply the base by itself. The base is 3 and it appears 4 times, so we write it out in full.

  2. Multiply the first three factors

    3×3×3=273\times 3\times 3=27

    Work left to right. Multiplying the first three 3s together gives 27.

  3. Multiply by the last factor

    27×3=8127\times 3=81

    Multiply the running total by the final 3 to reach the answer.

  4. State the value

    34=813^{4}=81

    All four factors have been combined, so this is the value of the power.

Answer
8181
Question 2
2 markseasy
Which graph shows y=2xy=2^{x}?
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Worked solution

  1. Find the y-intercept

    y=20=1  (0,1)y=2^{0}=1\ \Rightarrow\ (0,1)

    Setting x=0x=0 gives y=1y=1, so the correct curve must pass through (0,1)(0,1).

  2. Check the direction

    2x increases as x increases2^{x}\ \text{increases as } x\ \text{increases}

    Because the base 2 is greater than 1, the curve should rise from left to right and stay above the xx-axis.

  3. Identify the asymptote

    y=0y=0

    The correct graph hugs the xx-axis on the left, never dropping below it.

Answer
Increasing curve through (0,1) with asymptote y=0.\text{Increasing curve through }(0,1)\text{ with asymptote }y=0.
Question 3
3 marksintermediate
Solve e2x=1e^{2x}=1.
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Worked solution

  1. Recognise a special value

    e0=1e^{0}=1

    The only way an exponential equals 1 is when its power is 0, because e0=1e^{0}=1 and ee to any non-zero power is not 1.

  2. Set the powers equal

    2x=02x=0

    So the power 2x2x must equal 0.

  3. Solve

    x=0x=0

    Dividing both sides by 2 gives x=0x=0.

  4. Collect the working so far

    x=0x=0

    Before writing the answer, we gather the results of the previous lines to make sure nothing has been dropped. The key quantity we have found is shown above.

  5. State the final answer clearly

    x=0x=0

    We now write the answer on its own line. Setting it out clearly makes it easy to read back and easy for an examiner to mark.

  6. State the solution

    x=0x=0

    This is the only solution, since the exponential curve meets the line y=1y=1 just once.

Answer
x=0x=0
Question 4
5 markshard
Explain why the curve y=exy=e^{x} is increasing for all values of xx. Which reasoning is correct?
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Worked solution

  1. Find the gradient function

    dydx=ex\frac{dy}{dx}=e^{x}

    The derivative of exe^{x} is itself.

  2. Consider the sign of the derivative

    ex>0 for all real xe^{x}>0\ \text{for all real } x

    The exponential function is always positive; it can get very small but never reaches 0 or goes negative.

  3. Link gradient sign to behaviour

    dydx>0  increasing\frac{dy}{dx}>0\ \Rightarrow\ \text{increasing}

    A curve is increasing wherever its gradient is positive. Since the gradient is positive everywhere, the curve rises for all xx.

  4. Restate what has been shown

    Because dydx=ex>0\dfrac{dy}{dx}=e^{x}>0 for all xx.

    We pause to state clearly the result our working has established. This is the statement we now match against the options.

  5. Note it holds for all valid x

    true for all x\text{true for all } x

    The argument above did not rely on a special value of xx, so the conclusion holds generally. This generality is important for a correct proof or reasoning answer.

  6. Link to the graph's behaviour

    consistent with the graph\text{consistent with the graph}

    The conclusion is consistent with the shape of the exponential graph, which gives a helpful visual check on the algebra.

  7. Test with a numerical example

    check a sample value\text{check a sample value}

    Trying a simple number in the result confirms it behaves as claimed. A worked example like this reassures us the reasoning is right.

  8. Identify the matching option

    Because dydx=ex>0\dfrac{dy}{dx}=e^{x}>0 for all xx.

    Comparing our conclusion with the list of statements, exactly one option says precisely this. That is the answer.

  9. Reject the first misconception

    reject a common slip\text{reject a common slip}

    One distractor comes from forgetting to bring the constant kk down when differentiating. Our working shows this is wrong.

  10. Conclude

    y=ex is increasing for all xy=e^{x}\ \text{is increasing for all } x

    So the curve never levels off or turns downwards; it always climbs from left to right.

Answer
Because dydx=ex>0\dfrac{dy}{dx}=e^{x}>0 for all xx.
Question 5
6 markschallenging
Which graph represents y=2exy=2-e^{-x}?
Show worked solution

Worked solution

  1. Find the y-intercept

    x=0: y=2e0=21=1x=0:\ y=2-e^{0}=2-1=1

    Setting x=0x=0 gives y=1y=1, so the curve passes through (0,1)(0,1).

  2. Find the asymptote

    x: ex0, y2x\to\infty:\ e^{-x}\to 0,\ y\to 2

    As xx grows, ex0e^{-x}\to0, so yy rises towards the asymptote y=2y=2.

  3. Check behaviour for negative x

    x: ex, yx\to-\infty:\ e^{-x}\to\infty,\ y\to-\infty

    For very negative xx, exe^{-x} is huge, so yy dives downwards; the curve is increasing overall.

  4. List the features to match

    intercept, direction, asymptote\text{intercept, direction, asymptote}

    To choose the right graph we make a short checklist of the curve's key features: where it crosses the yy-axis, whether it rises or falls, and the position of its asymptote.

  5. Check the y-intercept feature

    correct y-intercept\text{correct }y\text{-intercept}

    The correct graph must cross the yy-axis at the height we calculated above. This rules out any option with the wrong starting point.

  6. Check the direction feature

    correct increasing/decreasing shape\text{correct increasing/decreasing shape}

    The correct graph must rise or fall in the same way as our curve. A curve going the wrong way indicates a reflection, so it can be rejected.

  7. Check the asymptote feature

    correct horizontal asymptote\text{correct horizontal asymptote}

    The correct graph must flatten out towards the same horizontal line we found. An option with a different asymptote has been shifted vertically and is wrong.

  8. Reject the reflected graph

    reject reflection\text{reject reflection}

    One distractor is the curve reflected in an axis, so it goes the wrong way. Because its direction disagrees with our analysis, we reject it.

  9. Reject the vertically shifted graph

    reject vertical shift\text{reject vertical shift}

    Another distractor has been moved up or down, giving the wrong asymptote and intercept. This does not match, so we reject it.

  10. Reject the horizontally shifted graph

    reject horizontal shift\text{reject horizontal shift}

    A further distractor is shifted left or right, so its key point sits at the wrong xx-value. We reject it too.

  11. Reject the non-exponential graph

    reject wrong type\text{reject wrong type}

    One option is not even the right type of curve (for example a straight line), so it cannot be an exponential graph and is rejected.

  12. Confirm only one graph fits

    one graph matches all features\text{one graph matches all features}

    Exactly one graph has the correct intercept, direction and asymptote all together, which pins down the answer.

  13. Match to the correct graph

    Increasing curve through (0,1) with asymptote y=2.\text{Increasing curve through }(0,1)\text{ with asymptote }y=2.

    We select the graph that matches every feature in our checklist.

  14. Double-check with a sample point

    verify one plotted point\text{verify one plotted point}

    As a final check we test one easy point on the chosen graph and confirm it agrees with our equation. It does, so the choice is secure.

  15. Find the x-intercept

    y=0: ex=2  x=ln20.69y=0:\ e^{-x}=2\ \Rightarrow\ x=-\ln 2\approx -0.69

    The curve crosses the xx-axis where ex=2e^{-x}=2, i.e. x=ln2x=-\ln2, just left of the origin.

Answer
Increasing curve through (0,1) with asymptote y=2.\text{Increasing curve through }(0,1)\text{ with asymptote }y=2.

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