# How do you tell whether f(x)=(6/5)^-x is an exponential growth or decay?

May 4, 2017

Given the form $f \left(x\right) = {A}^{-} x$

If $A > 1$ $\to$ exponential decay
If $A < 1$ $\to$ exponential growth

#### Explanation:

To understand why this makes sense, let's expand the function that is given. To do this, we need to keep in mind two properties of exponents:

First, ${\left(\frac{a}{b}\right)}^{x} = {a}^{x} / {b}^{x}$

Second, ${a}^{-} x = \frac{1}{a} ^ x$

Using the first property, expand the equation given in the problem:

$f \left(x\right) = {\left(\frac{6}{5}\right)}^{-} x = {6}^{-} \frac{x}{5} ^ - x$

Now, use the second property to flip the fraction and drop the negative sign from the exponent:

$f \left(x\right) = {5}^{x} / {6}^{x}$

Now that we've gotten this far, let's consider what will happen to the value of $f \left(x\right)$ as $x \to \infty$ by looking at the graphs of ${5}^{x}$ and ${6}^{x}$.

Graph of ${5}^{x}$
graph{5^x [-5, 5, -1.47, 5]}

Graph of ${6}^{x}$
graph{6^x[-5, 5, -1.47, 5]}

Looking at the difference between these two graphs, ${6}^{x}$ is steeper than ${5}^{x}$. This implies that the denominator of $f \left(x\right) = {5}^{x} / {6}^{x}$ will grow faster, making $f \left(x\right)$ smaller as $x$ increases (exponential decay).

If ${6}^{x}$ were in the numerator, the opposite would be true, and we would see exponential growth.

May 4, 2017

Decay

#### Explanation:

Given:$\text{ } f \left(x\right) = {\left(\frac{6}{5}\right)}^{- x}$

Set as:$y = {\left(\frac{6}{5}\right)}^{- x}$

The index (power) being negative is another way of writing:

$y = \frac{\textcolor{w h i t e}{.} 1 \textcolor{w h i t e}{.}}{\frac{6}{5}} ^ x$

This is the same as:

$y = {\left(\frac{5}{6}\right)}^{x} \text{ } = {5}^{x} / {6}^{x}$

As $x$ increases then ${6}^{x}$ becomes increasingly greater that ${5}^{x}$

Thus the whole will become less and less as $x$ becomes greater and greater eventually approaching a particular value.

Thus we have $y = {5}^{x} / {6}^{x} \to k$ as $x$ increases

And ${\lim}_{x \to \infty} y = {\lim}_{x \to \infty} k = 0$

Thus it is decay 