# How do you solve 36^(x^2+9) = 216^(x^2+9)?

##### 1 Answer
May 22, 2017

$x = \pm \sqrt{\frac{2 k \pi i}{\ln} 6 - 9}$

for any integer $k$

#### Explanation:

Given:

${36}^{{x}^{2} + 9} = {216}^{{x}^{2} + 9}$

Divide both sides by ${36}^{{x}^{2} + 9}$ to get:

$1 = {6}^{{x}^{2} + 9} = {e}^{\left({x}^{2} + 9\right) \ln 6}$

From Euler's identity we can deduce:

$\left({x}^{2} + 9\right) \ln 6 = 2 k \pi i$

for any integer $k$

So:

${x}^{2} + 9 = \frac{2 k \pi i}{\ln} 6$

Hence:

$x = \pm \sqrt{\frac{2 k \pi i}{\ln} 6 - 9}$

If $k = 0$ that gives us:

$x = \pm \sqrt{- 9} = \pm 3 i$

$\textcolor{w h i t e}{}$
Footnote

If you would like the other roots in $a + b i$ form, then you can use the formula derived in https://socratic.org/s/aEUsUcjD , namely that the square roots of $a + b i$ are:

$\pm \left(\left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2}}\right) + \left(\frac{b}{\left\mid b \right\mid} \sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}\right) i\right)$