# How do you solve 12(1-4^x)=18?

Jul 31, 2016

$x = - \frac{1}{2} + \frac{\left(2 k + 1\right) \pi i}{\ln} \left(4\right)$ for any integer $k$

#### Explanation:

Divide both sides by $12$ to get:

$1 - {4}^{x} = \frac{18}{12} = \frac{3}{2}$

Add ${4}^{x} - \frac{3}{2}$ to both sides to get:

$- \frac{1}{2} = {4}^{x}$

For any Real value of $x$, ${4}^{x} > 0$, so it cannot equal $- \frac{1}{2}$.

If we had wanted ${4}^{x} = \frac{1}{2}$ then we could have found:

${2}^{- 1} = \frac{1}{2} = {4}^{x} = {\left({2}^{2}\right)}^{x} = {2}^{2 x}$

and hence solution $x = - \frac{1}{2}$.

We can make this into a set of Complex solutions for our original problem by adding an odd multiple of $\frac{\pi i}{\ln} \left(4\right)$

Let $x = - \frac{1}{2} + \frac{\left(2 k + 1\right) \pi i}{\ln} \left(4\right)$ for any integer $k$

Then:

${4}^{x} = {4}^{- \frac{1}{2} + \frac{\left(2 k + 1\right) \pi i}{\ln} \left(4\right)}$

$= {4}^{- \frac{1}{2}} \cdot {4}^{\frac{\left(2 k + 1\right) \pi i}{\ln} \left(4\right)}$

$= \frac{1}{2} \cdot {\left({e}^{\ln} \left(4\right)\right)}^{\frac{\left(2 k + 1\right) \pi i}{\ln} \left(4\right)}$

$= \frac{1}{2} \cdot {e}^{\ln \left(4\right) \cdot \frac{\left(2 k + 1\right) \pi i}{\ln} \left(4\right)}$

$= \frac{1}{2} \cdot {e}^{\left(2 k + 1\right) \pi i}$

$= \frac{1}{2} \cdot {\left({e}^{\pi i}\right)}^{2 k + 1}$

$= \frac{1}{2} \cdot {\left(- 1\right)}^{2 k + 1}$

$= - \frac{1}{2}$