Model practical problems with exponential functions of the form $y = a b^x$ and interpret growth, decay and asymptotes

What this dot point is asking

NESA wants you to identify when an exponential model applies (anything where the rate of change is proportional to the current size: bacterial growth, radioactive decay, compound interest, depreciation), set up $y = a b^x$ with the right base, evaluate it, and solve exponential equations using logarithms.

The answer

The standard form

• IMATH_6 is the initial value at $x = 0$ (since $b^0 = 1$).
• IMATH_9 is the base, the per-unit multiplier.
• IMATH_10 : growth. $0 < b < 1$: decay.

From percentage rate to base IMATH_12

• Growth at $r\%$ per period: $b = 1 + \frac{r}{100}$. So $7\%$ growth gives $b = 1.07$.
• Decay at $r\%$ per period: $b = 1 - \frac{r}{100}$. So $12\%$ decay gives $b = 0.88$.

This is the same multiplier as the one used in straight compound interest or declining-balance depreciation.

Asymptote

The graph of $y = a b^x$ (for $a > 0$) approaches the $x$-axis but never touches it. The line $y = 0$ is a horizontal asymptote. Practical interpretation: a decaying quantity gets closer and closer to zero without ever reaching exactly zero.

Solving for the exponent

If $y = a b^x$ and you know $y$, $a$ and $b$ and want $x$:

Either log base works because the bases cancel in the ratio. Use whichever your calculator gives easily.

Reading off worded problems

• "Doubles every $T$ time units" implies $b = 2$ per $T$, so per unit of time the base is $2^{1/T}$.
• "Half-life of $T$" implies $b = 0.5$ per $T$, so per unit $b = 0.5^{1/T}$.
• "Grows at $r\%$ continuously" is technically $e^{r t}$, but Standard 2 uses discrete compounding, so use $b = 1 + r/100$.