Exponential Function Manipulation — AP® Precalculus Notes

The same exponential function can wear many outfits. $2^{x + 3}$ , $8(2)^x$ and $4(2)^{x + 1}$ are one function. $5(4)^x$ and $5(2)^{2x}$ are one function. The exponent rules from "Algebra you need" turn one form into another, and each form has a graphical meaning. This page shows both, because the exam asks you to rewrite exponential expressions exactly and to explain what the rewriting says about the graph.

The rules you use

Rule	Rewrite	What it says about the graph
Product: $b^{m + n} = b^m b^n$	$b^{x + k} = b^k \cdot b^x$	A horizontal translation is a vertical dilation by $b^k$
Power: $b^{mn} = (b^m)^n$	$b^{cx} = (b^c)^x$	A horizontal dilation is a change of base
Negative: $b^{-n} = \frac{1}{b^n}$	$b^{-x} = \left(\frac{1}{b}\right)^x$	Reflecting growth across the $y$ -axis gives decay
Rational: $b^{1/n} = \sqrt[n]{b}$	$b^{x/n} = (\sqrt[n]{b})^x$	A horizontal stretch by $n$ takes the $n$ th root of the base

A horizontal shift is a vertical stretch

For most functions, moving the graph sideways and stretching it upward are different things. For an exponential function they are the same.

Worked example 1. Show that $h(x) = 2^{x + 3}$ is a vertical dilation of $y = 2^x$ .

The exponent is a sum, so use the product rule backward: $2^{x + 3} = 2^x \cdot 2^3$ .

$2^3 = 8$ , so $h(x) = 8(2)^x$ . Every output of $2^x$ is multiplied by $8$ .

Check one point: $h(0) = 2^3 = 8$ and $8(2)^0 = 8$ .

The graph of y = 2^x through (0, 1) and the graph of h(x) = 2^(x + 3) through (-3, 1) and (0, 8): the second is the first moved 3 units left, and also the first with every output multiplied by 8. — **Figure 1.** $2^{x + 3}$ : three units left, or eight times as tall. Both descriptions are right. LevlPrep original diagram.

In the figure, the point $(0, 1)$ on $2^x$ moves to $(-3, 1)$ under the shift, and the point $(0, 1)$ also lines up with $(0, 8)$ under the stretch. Shifting right instead divides: $5(4)^{x - 1} = \frac{5}{4}(4)^x$ .

The trap is reading $2^{x + 3}$ as $2^x + 2^3$ . Adding in the exponent means multiplying the powers. Test it: at $x = 1$ , $2^4 = 16$ but $2^1 + 8 = 10$ .

A horizontal dilation is a new base

Worked example 2. Rewrite $g(x) = 5(4)^x$ with base $2$ , and $2^{3x}$ with a single base.

$4 = 2^2$ , so $4^x = (2^2)^x = 2^{2x}$ . Then $g(x) = 5(2)^{2x}$ : the graph of $5(2)^x$ compressed horizontally by a factor of $\frac{1}{2}$ .

$2^{3x} = (2^3)^x = 8^x$ : compressing $2^x$ by a factor of $\frac{1}{3}$ gives base $8$ .

It works with fractions too. $25^{x/2}$ $= (25^{1/2})^x$ $= 5^x$ , and $\left(\frac{1}{3}\right)^x$ $= (3^{-1})^x$ $= 3^{-x}$ .

Rewriting with every step shown

Rewrite each base as a power of a common base, apply the power rule to the whole exponent, then combine.

Worked example 3. Write $\frac{4^{x + 1}}{2^x}$ as a single power of $2$ .

Common base: $4^{x + 1} = (2^2)^{x + 1}$ .

Power rule on the whole exponent: $(2^2)^{x + 1}$ $= 2^{2(x + 1)}$ $= 2^{2x + 2}$ . (Writing $2^{2x + 1}$ forgets to double the $1$ .)

Divide, same base, subtract exponents: $\frac{2^{2x + 2}}{2^x}$ $= 2^{2x + 2 - x}$ $= 2^{x + 2}$ .

These rewrites also let you use a known value. If $3^x = 7$ , then $3^{x + 2} = 3^x \cdot 3^2 = 63$ , and $3^{2x} = (3^x)^2 = 49$ .

On the exam, a "single power" answer means one base raised to one exponent, with no sums or quotients left over. Stopping at $\frac{2^{2x + 2}}{2^x}$ is not finished.