A linear change adds or removes the same fixed amount every period. An exponential change instead applies the same fixed percentage of the current value every period — so the actual amount added or removed keeps changing as the value itself changes. Populations, investments, radioactive substances, and depreciating vehicles all typically behave this way.

where $N_0$ is the initial value, $r\%$ is the rate per time period, $t$ is the number of periods, $+$ is used for growth and $-$ for decay.

Identify $N_0$, $r$, and $t$, substitute into the formula, and evaluate the power before multiplying by $N_0$.

Worked Example

A machine is bought for \18\,000$. It depreciates exponentially at$12\%$per year. Find its value after$6$ years, correct to the nearest dollar.

1. Decay multiplier: $1-\dfrac{12}{100}=0.88$.
2. Apply the formula: $N = 18\,000 \times (0.88)^6$.
3. Evaluate: $(0.88)^6 = 0.4644\ldots$, so $N = 18\,000\times0.4644\ldots = 8359.4\ldots$
4. Round: N \approx \8359$.

Nothing changes except the multiplier — use $1+\dfrac{r}{100}$ instead of $1-\dfrac{r}{100}$, since the quantity is increasing by a fixed percentage of itself each period rather than decreasing.

Worked Example

A bacteria population starts at $2000$ and grows exponentially at a rate of $8\%$ per hour. Find the population after $5$ hours, correct to the nearest whole number.

1. Growth multiplier: $1+\dfrac{8}{100}=1.08$.
2. Apply the formula: $N = 2000\times(1.08)^5$.
3. Evaluate: $(1.08)^5 = 1.46932\ldots$, so $N = 2000\times1.46932\ldots = 2938.6\ldots$
4. Round to a whole number of bacteria: $N\approx2939$.

If $t$ is unknown, you would need logarithms to solve directly — not required at this level, so such problems are normally restricted to finding $N_0$, $r$, or $N$ given the other quantities, or to comparing two given scenarios (e.g. "after how many whole years does the value first drop below \X$?") by substituting increasing whole-number values of$t$ and checking.