Introduction

In this unit, we explore how to rewrite exponential expressions in equivalent forms. 

The Product Property

$b^{m+n}=b^m\cdot b^n$

The first major concept related horizontal shifts to vertical stretches. It is the Product Property.

When we have an exponential function with a horizontal translation (a shift left or right), it takes the shape of $f\left(x\right)=b^{x+k}$. Here, $+k$ represents a shift left by $k$ units.

Note, you may be used to the form $f\left(x\right)=b^{x-k}$. This used to be the standard formula Collegeboard used. However, they recently changed the minus sign into a plus sign. This doesn’t have any effect on the function itself, though, it will have an effect on $k$. For example, if you were asked about the value of $k$, and you had the function $f\left(x\right)=2^{x-3}$, using the old formula ($f\left(x\right)=b^{x-k}$), you would give the answer $k=3$. However, using the new formula ($f\left(x\right)=b^{x+k}$), you would say $k=-3$. 

To reiterate, there is NO effect on the actual function’s behavior. They are the same function, we just redefined how we choose our $k$ value.

However, using the product property, we can rewrite this as $f\left(x\right)=b^k{b}^x$. Because $b^k$ is a constant number, we can rewrite it as a constant, which we’ll call $a$. This turns the function into $f\left(x\right)=a{b}^x$. 

Graphically, this rule implies that every horizontal translation of an exponential function is equivalent to some vertical dilation (stretch/compression).

The Division Property

The second rule is an extension to the product property. It states that 

$b^{m-n}=\frac{b^m}{b^n}$

Graphically, this also implies that every horizontal translation is equivalent to some vertical dilation. 

The Power Property

The third rule relates horizontal stretches/compressions to changing the base. 

$(b^m{)}^n=b^{mn}$

What this rule is saying is that when we have an exponential function with a horizontal dilation ($f\left(x\right)=b^{cx}$), we can use the power property to group the base $b$ with the coefficient $c$ (note, we are using the reverse of this property: $b^{mn}=(b^m{)}^n$). This turns into $f\left(x\right)={\left(b^c\right)}^x$. Then, we can calculate the new value of $b^c$ to find a new base (which we’ll call $B$). Substituting $B$ back, we have $f\left(x\right)={\left(B\right)}^x$

This implies that for every horizontal dilation of an exponential function, there is an equivalent form of an exponential function with a different base: 

Negative Exponents

The Negative Exponent Property states that 

$b^{-n}=\frac{1}{b^n}={\left(\frac{1}{b}\right)}^n$

This is useful for simplifying exponential expressions. However, there is another, more useful way to use this property: to change the base of an exponential expression.

Exponential Growth/Decay

Exponential growth occurs when the function’s value increases at an increasing rate. Essentially, as $x$ increases, the $y$-values grow larger and faster. For an exponent function $f(x)=a{b}^x$ to exhibit exponential growth, $b>1$.

In limit notation, the following statement can be translated as 

For Exponential decay, the function’s output values decrease at a decreasing rate. For an exponent function $f(x)=a{b}^x$ to exhibit exponential decay, $b<1$. If $b=1$, the exponent function turns to a constant function. Note that sometimes, exponential decay is written as $f(x)=a{b}^{-x}$, where $a>0$ and $b>0$. 

In limit notation, the following statement can be translated as 

Both exponential growth and decay are functions that concave up. 

Unit Fractions

This rule explains how to evaluate fractional exponents. It says:

$(b{)}^{}{}^{\frac{1}{k}}=\sqrt[k]{b}$.

Given that $\sqrt[k]{b}$ is defined. For example, $(-1{)}^{\frac{1}{2}}$ does not exist because $\sqrt{-1}$ does not exist.

Graphically, this implies that if the input is scaled by a fraction, it is a horizontal stretch, which is equivalent to taking the square root to the base. This rule is primarily used for simplification purposes.

Practice Problems

1. Let $f\left(x\right)=3^x$. Describe the function $g\left(x\right)=81\cdot 3^x$ in terms of $f\left(x\right)$ and a translation.

2. Let $f\left(x\right)=2^x$. Describe the function $g\left(x\right)=32^x$ in terms of $f\left(x\right)$ and a dilation.

3. Simplify the following (without a calculator). You may leave your answer in the form $b^n$ where $b$ is a prime number${\left(9^{2.1}\right)}^3\cdot \sqrt[4]{9^{6.8}}$.

Answer Key

#1:

$81=3^4$, so

$g\left(x\right)=3^4\cdot 3^x=3^{x+4}$

Since $f\left(x\right)=3^x$, $g\left(x\right)=f\left(x+4\right)$, which corresponds to a horizontal translation left of 4 units.

#2

$32=2^5$

$g\left(x\right)={\left(2^5\right)}^x$

$g\left(x\right)=\left(2^{5x}\right)$

Since $f\left(x\right)=2^x$, $g\left(x\right)=f\left(5x\right)$, which corresponds to a horizontal dilation by 5.

#3

${\left(9^{2.1}\right)}^3\cdot \sqrt[4]{9^{6.8}}$

$9^{6.3}\cdot \sqrt[4]{9^{6.8}}$

$9^{6.3}\cdot 9^{\frac{6.8}{4}}$

$9^{6.3}\cdot 9^{1.7}$

$9^8$

${\left(3^2\right)}^8$

$3^{16}$