An Exponential Growth and Decay Problem

Exponential models put $e^x$ and $\ln$ into a real context, temperature, population, value. The two tasks are always the same: substitute a time to find a value, and take logs to find a time. The method mirrors any exponential equation solved with ln.

The value $V$ of an investment, in ringgit, after $t$ years is modelled by $V = 5000e^{0.02t}$. (i) Find the value after $10$ years. [2] (ii) Find, to the nearest year, when the value first reaches $8000$. [4]

The working

(i) Substitute $t = 10$: $V = 5000e^{0.02 \times 10} = 5000e^{0.2} = 5000(1.2214\ldots) = 6107 \quad \text{(M1, A1)}$

So the value after $10$ years is about RM6107.

(ii) Set $V = 8000$ and isolate the exponential: $8000 = 5000e^{0.02t} \;\Rightarrow\; e^{0.02t} = \frac{8000}{5000} = 1.6 \quad \text{(M1)}$

Take natural logs (the base is $e$): $0.02t = \ln 1.6 = 0.4700\ldots \quad \text{(M1)}$ $t = \frac{0.4700}{0.02} = 23.5 \quad \text{(A1)}$

To the nearest year, the value first reaches $8000$ after $24$ years (round up, since at $t = 23$ it hasn’t quite reached $8000$). (A1)

Where the marks are won and lost

• Divide before you log. Isolate $e^{0.02t}$ first; taking $\ln$ of the whole equation without dividing is messier and error-prone.
• Use $\ln$, not $\lg$, because the base is $e$: $\ln(e^{0.02t}) = 0.02t$ cleanly.
• “First reaches” plus “nearest year” means round up to $24$: at $23$ years the target isn’t met yet, so $24$ is the first whole year it’s reached.

Common mistakes

• Using $\log$ base 10 instead of $\ln$.
• Rounding $\ln 1.6$ too early and drifting off the answer.
• Rounding $23.5$ down to $23$ when “first reaches” requires rounding up.

Full method: e^x and ln x notes. Topic home: Logs & Exponentials pillar.