A model y=ab^t passes through (0,300) and (5,600). Find b to 4 decimal places.

Write the model: y = 300b^t Use the second point: 600 = 300b^5 Divide by 300: b^5 = 2 Take the 5th root: b = 2^1/5 Evaluate: b = 1.1487 Write the full model: y = 300× 1.1487^ t Answer: b = 1.1487

A-Level Exponential modelling Practice Questions

Free A-Level Exponential modelling practice questions with full step-by-step worked solutions. Covers growth model, initial value, substitution, decay model. Practise exam-style problems and check your method.

growth modelinitial valuesubstitutiondecay modely=ab^tsolving

A-Level70 questionsStep-by-step solutions

Question 1

2 markseasy

A population of insects is modelled by

P=500e^{0.03t}

, where

t

is the time in years. State the initial population.

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Worked solution

Write down the model
$P = 500e^{0.03t}$
This models a population P after t years. The number in front, 500, sits outside the exponential.
Substitute t = 0
$P = 500e^{0.03 \times 0}$
The initial population means the value at the very start, when time t is zero.
Simplify the power
$P = 500e^{0} = 500 \times 1$
Anything to the power 0 equals 1, so e^{0}=1. This is a key index law from earlier work on indices.
State the initial value
$P = 500$
So the model starts at 500. The constant A in P=Ae^{kt} always gives the starting amount.

Answer

500

Question 2

2 markseasy

When

\ln y

is plotted against

t

for the model

y=Ae^{kt}

, a straight line of gradient

0.25

is obtained. State the value of

k

Show worked solution

Worked solution

Start from the model
$y = Ae^{kt}$
We are told that plotting ln y against t gives a straight line. Let us see why.
Take natural logs of both sides
$\ln y = \ln A + kt$
Using log laws, the product becomes a sum and the power kt comes down. This is a linearising trick used a lot in this topic.
Compare with y = mx + c
$\ln y = (k)\,t + \ln A$
Reading it as a straight line, t is the horizontal variable, so the gradient is k and the intercept is ln A.
State k
$k = \text{gradient} = 0.25$
The gradient of the ln y against t line is exactly the growth constant k.

Answer

k = 0.25

Question 3

4 marksintermediate

A model

y=ab^{t}

passes through

(0,300)

and

(5,600)

. Find

b

4

decimal places.

Show worked solution

Worked solution

Write the model
$y = 300b^{t}$
At t=0 the value is 300, which gives the coefficient. We still need the base b.
Use the second point
$600 = 300b^{5}$
At t=5 the value is 600, giving an equation for b.
Divide by 300
$b^{5} = 2$
So over 5 units of time the quantity doubles.
Take the 5th root
$b = 2^{1/5}$
To undo the power of 5 we raise both sides to the power 1/5, using index laws.
Evaluate
$b = 1.1487$
So b ≈ 1.1487, meaning about a 14.87% increase each time step.
Write the full model
$y = 300\times 1.1487^{\,t}$
Combining a=300 and b gives the complete model. This matches the two data points we were given.

Answer

b = 1.1487

Question 4

4 markshard

Which graph represents

y=6b^{x}

where

0<b<1

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Worked solution

Find the y-intercept
$x=0\Rightarrow y=6b^{0}=6$
Every model y=ab^x passes through (0,a) because b^0=1. Here that point is (0,6).
Decide growth or decay
$0<b<1\Rightarrow\text{decay}$
When the base b is between 0 and 1, repeatedly multiplying makes the value smaller, so the curve decreases.
Behaviour as x increases
$x\to\infty\Rightarrow y\to 0$
As x grows, b^x shrinks towards 0, so the curve approaches the x-axis from above.
Behaviour as x decreases
$x\to-\infty\Rightarrow y\to\infty$
For negative x, b^x becomes large, so the curve rises steeply on the left.
Note the curve stays positive
$6b^{x}>0\ \text{for all }x$
Since 6 and b^x are both positive, y is never negative; the curve stays above the x-axis.
Rule out the increasing curve
$\text{rising} \Rightarrow b>1$
A curve that grows would need a base bigger than 1, contradicting 0<b<1.
Rule out the reflected curve
$\text{below the axis} \Rightarrow \text{negative }a$
A curve dipping below the x-axis would need a negative coefficient, but here a=6>0.
Rule out the shifted curve
$\text{levels off above }0 \Rightarrow \text{extra constant}$
A curve flattening towards y=2 would be 6b^x+2, but our model has no added constant.
Rule out the straight line
$\text{constant rate} \Rightarrow \text{not exponential}$
Decay slows as y gets smaller, so the graph must curve rather than fall in a straight line.
Choose the correct graph
$\text{curve through }(0,6)\text{ decreasing to }y=0$
So the correct graph starts at (0,6) and decays towards the x-axis, always staying positive.

Answer

Positive decreasing curve through

(0,6)

approaching

y=0

Question 5

7 markschallenging

Light intensity in a lake is

I=I_0e^{-0.15x}

, where

x

is depth in metres. Find, to

2

decimal places, the depth at which the intensity is

5\%

of the surface value.

Show worked solution

Worked solution

Write the model
$I = I_0e^{-0.15x}$
Light intensity falls with depth x metres; μ=0.15 per metre is the attenuation constant.
Set up the 5% equation
$e^{-0.15x} = 0.05$
We want the depth where intensity is 5% of the surface value; the I_0 cancels.
Take natural logs
$-0.15x = \ln 0.05$
Logs bring the power down.
Evaluate ln 0.05
$-0.15x = -2.9957$
ln 0.05 ≈ -2.9957 (negative, since 0.05 < 1).
Solve for the depth x
$x = \dfrac{-2.9957}{-0.15} = 19.97$
So the intensity is 5% at about 19.97 m depth.
Now find the intensity at 10 m
$\dfrac{I}{I_0} = e^{-0.15\times 10}$
Use the same model to check the intensity at 10 m.
Work out the power
$-0.15\times 10 = -1.5$
Multiply out the exponent.
Evaluate
$e^{-1.5} = 0.2231$
e^{-1.5} ≈ 0.2231.
Convert to a percentage
$22.31\%$
So about 22.31% of surface light remains at 10 m. The required answer is the depth, ≈ 19.97 m.
State the final answer clearly
$x = 19.97\ \text{m}$
Here is our result written out on its own. It is worth stating it clearly, with the correct units and to the accuracy the question asked for.
Interpret the answer in context
$x = 19.97\ \text{m}$
Always turn the number back into words for the real situation. Ask yourself: does this value make sense for the population, temperature, mass or money in the question?
Why logarithms were the key tool
$\ln\left(e^{x}\right)=x,\qquad \log\left(10^{x}\right)=x$
The variable started trapped inside a power. Logs work because they are the exact inverse of raising a base to a power, so they bring the unknown down where we can reach it.
Link back to the log and index laws
$\log(pq)=\log p+\log q,\quad \log\left(p^{n}\right)=n\log p$
These are the same log and index rules you met earlier in the course. Exponential modelling simply uses them in a real-world setting, so revising them always helps here.
Watch out for a common error
$\text{isolate the exponential BEFORE taking logs}$
A very common slip is to take logs while a number is still multiplying the exponential term. Always get the $e^{\dots}$ (or the power) on its own first, then log.
Check the answer is reasonable
$x = 19.97\ \text{m}$
Sanity-check the size and sign: a growth model should give increasing values and a decay model decreasing ones, and a time should come out positive.

Answer

x = 19.97\ \text{m}

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