# What is the domain and range of (2/3)^x – 9?

May 28, 2017

Domain: $\left(- \infty , \infty\right)$
Range: $\left(- 9 , \infty\right)$

#### Explanation:

First note that ${\left(\frac{2}{3}\right)}^{x} - 9$ is well defined for any Real value of $x$. So the domain is the whole of $\mathbb{R}$, i.e. $\left(- \infty , \infty\right)$

Since $0 < \frac{2}{3} < 1$, the function ${\left(\frac{2}{3}\right)}^{x}$ is an exponentially decreasing function which takes large positive values when $x$ is large and negative, and is asymptotic to $0$ for large positive values of $x$.

In limit notation, we can write:

${\lim}_{x \to - \infty} {\left(\frac{2}{3}\right)}^{x} = - \infty$

${\lim}_{x \to \infty} {\left(\frac{2}{3}\right)}^{x} = 0$

${\left(\frac{2}{3}\right)}^{x}$ is continuous and strictly monotonically decreasing, so its range is $\left(0 , \infty\right)$.

Subtract $9$ to find that the range of ${\left(\frac{2}{3}\right)}^{x}$ is $\left(- 9 , \infty\right)$.

Let:

$y = {\left(\frac{2}{3}\right)}^{x} - 9$

Then:

$y + 9 = {\left(\frac{2}{3}\right)}^{x}$

If $y > - 9$ then we can take logs of both sides to find:

$\log \left(y + 9\right) = \log \left({\left(\frac{2}{3}\right)}^{x}\right) = x \log \left(\frac{2}{3}\right)$

and hence:

$x = \log \frac{y + 9}{\log} \left(\frac{2}{3}\right)$

So for any $y \in \left(- 9 , \infty\right)$ we can find a corresponding $x$ such that:

${\left(\frac{2}{3}\right)}^{x} - 9 = y$

That confirms that the range is the whole of $\left(- 9 , \infty\right)$.