# How do you solve (1/3)^x = -3 ?

Dec 8, 2016

$x = - 1 + \frac{\left(2 k + 1\right) \pi}{\ln \left(3\right)} i \text{ }$ for any integer $k$

#### Explanation:

For any Real value of $x$ we have ${\left(\frac{1}{3}\right)}^{x} > 0$. So there is no Real solution.

There are Complex solutions. Consider Euler's identity:

${e}^{i \pi} = - 1$

Hence:

${e}^{i \left(2 k + 1\right) \pi} = - 1 \text{ }$ for any integer $k$

Now:

${\left(\frac{1}{3}\right)}^{x} = {\left({e}^{\ln \left(\frac{1}{3}\right)}\right)}^{x} = {e}^{- x \ln \left(3\right)}$

So if $\text{ "x = -1 + ((2k+1)pi)/(ln(3)) i" }$ then we find:

${\left(\frac{1}{3}\right)}^{x} = {e}^{- \ln \left(3\right) \left(- 1 + \frac{\left(2 k + 1\right) \pi}{\ln \left(3\right)} i\right)}$

$\textcolor{w h i t e}{{\left(\frac{1}{3}\right)}^{x}} = {e}^{\ln \left(3\right) + \left(2 k + 1\right) \pi i}$

$\textcolor{w h i t e}{{\left(\frac{1}{3}\right)}^{x}} = {e}^{\ln \left(3\right)} \cdot {e}^{\left(2 k + 1\right) \pi i}$

$\textcolor{w h i t e}{{\left(\frac{1}{3}\right)}^{x}} = 3 \cdot \left(- 1\right)$

$\textcolor{w h i t e}{{\left(\frac{1}{3}\right)}^{x}} = - 3$