# How do you solve 2^ { 2x } - ( 2^ { x + 2} ) - 32= 0?

Apr 29, 2017

$x = 3$

#### Explanation:

First of all, you should know that any number of the form ${a}^{b c}$ can be written in the form ${\left({a}^{b}\right)}^{c}$ or ${\left({a}^{c}\right)}^{b}$ or vice-versa.

i.e. ${a}^{b c} = {\left({a}^{b}\right)}^{c} = {\left({a}^{c}\right)}^{b}$.

Also, ${a}^{b + c} = {a}^{b} \cdot {a}^{c}$

$\therefore$ above equation ${2}^{2 x} - \left({2}^{x + 2}\right) - 32 = 0$ can be writte as:-

${\left({2}^{x}\right)}^{2} - {2}^{2} \cdot {2}^{x} - 32 = 0$

$\implies {\left({2}^{x}\right)}^{2} - 4 \cdot {2}^{x} - 32 = 0$

Let ${2}^{x}$ be some number $y$.

$\therefore {y}^{2} - 4 y - 32 = 0$

Now solving it just like a regular quadratic equation

$\implies {y}^{2} - 8 y + 4 y - 32 = 0$

$\implies y \left(y - 8\right) + 4 \left(y - 8\right) = 0$

$\implies \left(y + 4\right) \left(y - 8\right) = 0$

$y = - 4 , 8$

But, $y = {2}^{x}$

$\implies {2}^{x} = - 4 , 8$

Since any exponential function defined on real numbers cannot have a negative value, we will discard $- 4$.

graph{2^x [-10, 10, -5, 5]}

You can observe in this graph of ${2}^{x}$ that no matter what the value of $x$, the value of $y = {2}^{x}$ is always positive. Its value is getting closer and closer to zero but is never actually touching zero. This is true for any function of the type ${a}^{x}$ where $a$ and $x$ are REAL Numbers.

$\therefore$ ${2}^{x} = 8$

$\implies {2}^{x} = {2}^{3}$

$\implies x = 3$