How do you solve 3^(x^2 + 20) = (1/27)^(3x)?

Aug 20, 2015

The solutions are
color(blue)(x=-4,x=-5

Explanation:

${3}^{{x}^{2} + 20} = {\left(\frac{1}{27}\right)}^{3 x}$

We know that 1/27=1/(3^3

So,

${3}^{{x}^{2} + 20} = {\left(\frac{1}{3} ^ 3\right)}^{3 x}$

By property
color(blue)(1/a=a^-1

${3}^{{x}^{2} + 20} = {\textcolor{b l u e}{\left({3}^{-} 3\right)}}^{3 x}$

${3}^{{x}^{2} + 20} = {3}^{- 9 x}$
Now as bases are equal we equate powers and find $x$

${x}^{2} + 20 = - 9 x$

${x}^{2} + 9 x + 20 = 0$

We can Split the Middle Term of this expression to factorise it and find solutions.

${x}^{2} + \textcolor{b l u e}{9 x} + 20 = 0$

${x}^{2} + \textcolor{b l u e}{5 x + 4 x} + 20 = 0$

$x \left(x + 5\right) + 4 \left(x + 5\right) = 0$

$\left(x + 4\right) \left(x + 5\right) = 0$

We now equate the factors to zero.
$x + 4 = 0 , x = - 4$

$x + 5 = 0 , x = - 5$