# How do you solve [(2^(2x)) - 3] * [(2^(x+2)) + 32] = 0?

Apr 11, 2018

$x \approx .7925$

#### Explanation:

We can start by dividing one term from both side of the equation and simplifiying:

$\left[\left({2}^{2 x} - 3\right)\right] \cdot \left[\left({2}^{x + 2}\right) + 32\right] = 0$

$\left[\left({2}^{2 x} - 3\right)\right] \cdot \frac{\left({2}^{x + 2}\right) + 32}{\left({2}^{x + 2}\right) + 32} = \frac{0}{\left({2}^{x + 2}\right) + 32}$

${2}^{2 x} - 3 = 0$

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

Now we can take the Log of the equation with the base being that of the number with the variable in the exponent:

${\log}_{2} {2}^{2 x} = {\log}_{2} 3$

As you may remember, this is equivalent to:

$2 x \cdot \left(1\right) = 1.585 \ldots$

Finally, solve for $x$

$x \approx .7925$