# How do you solve (log(x))^2=4?

Jun 14, 2016

$x = {10}^{2}$ or $x = {10}^{-} 2$

#### Explanation:

${\left(L o g \left(x\right)\right)}^{2} = 4$
$\implies {\left(L o g \left(x\right)\right)}^{2} - {2}^{2} = 0$

Use formula named as Difference of Squares which states that if ${a}^{2} - {b}^{2} = 0$, then $\left(a - b\right) \left(a + b\right) = 0$

Here ${a}^{2} = {\left(L o g \left(x\right)\right)}^{2}$ and ${b}^{2} = {2}^{2}$

$\implies \left(\log \left(x\right) - 2\right) \left(\log \left(x\right) + 2\right) = 0$

Now, use Zero Product Property which states that if the product of two number, say $a$ and $b$, is zero then one of two must be zero, i.e., either $a = 0$ or $b = 0$.

Here $a = \log \left(x\right) - 2$ and $b = \log \left(x\right) + 2$

$\implies$ either $\log \left(x\right) - 2 = 0$ or $\log \left(x\right) + 2 = 0$
$\implies$ either $\log \left(x\right) = 2$ or $\log \left(x\right) = - 2$
$\implies$ either $x = {10}^{2}$ or $x = {10}^{-} 2$