Linear functions add the same amount at every step. Many things in the world multiply instead: a population that grows by half each decade, money earning interest, a medicine leaving the body. Those follow exponential functions, which use the exponents you know with the variable in the exponent. With function notation we write $f(x) = a \cdot b^x$.
Why it works
Start at $f(0) = a \cdot b^0 = a$, since $b^0 = 1$. Each time $x$ goes up by $1$, one more factor of $b$ appears: $f(x + 1) = a \cdot b^{x+1} = (a \cdot b^x) \cdot b = f(x) \cdot b$. So $a$ is the starting value and $b$ is the factor that every step multiplies by. If $b > 1$ the outputs grow; if $0 < b < 1$ they shrink (decay). And $b$ must be positive, so that $b^x$ makes sense for every $x$.
Key idea
$f(x) = a \cdot b^x$ with $b > 0$, $b \ne 1$: the starting value is $a$, and each step of $1$ in $x$ multiplies the output by $b$.
Worked examples
Worked example: Evaluate
For $f(x) = 3 \cdot 2^x$, find $f(0)$, $f(4)$ and $f(-1)$.
- $f(0) = 3 \cdot 1 = 3$.
- $f(4) = 3 \cdot 2^4 = 3 \cdot 16 = 48$. The power comes before the multiplication.
- $f(-1) = 3 \cdot 2^{-1} = 3 \cdot \frac{1}{2} = \frac{3}{2}$.
Answer: $3$, $48$ and $\frac{3}{2}$.
Worked example: Decay
For $g(x) = 80\left(\frac{1}{2}\right)^x$, find $g(3)$ and describe $g$.
- $g(3) = 80 \cdot \frac{1}{8} = 10$.
- The base $\frac{1}{2}$ is between $0$ and $1$: each step halves the output. Starting value $80$.
Answer: $g(3) = 10$; exponential decay that halves at each step.
Common mistake
Two classic slips: mixing up $2^x$ with $x^2$, and multiplying the coefficient into the base. $2^x$ has the variable in the exponent ($2^5 = 32$), while $x^2$ has it in the base ($5^2 = 25$). And $3 \cdot 2^x$ is not $6^x$: at $x = 2$ it is $3 \cdot 4 = 12$, while $6^2 = 36$. Order of operations puts the power first.