# How do you solve 0.5^x=16^2?

Sep 8, 2016

$x = - 8$

#### Explanation:

When you are working with equations with indices, try to

either $\rightarrow \text{ make the bases the same}$

or $\rightarrow \text{ make the indices the same}$

$0.5 = \frac{1}{2}$ This is a better form to use, because you should recognise that 16 is one of powers of 2.......

$\rightarrow {2}^{4} = 16 \mathmr{and} {2}^{8} = {16}^{2}$

${\left(\frac{1}{2}\right)}^{x} = {16}^{2}$

${\left(\frac{1}{2}\right)}^{x} = {2}^{8}$

${2}^{-} x = {2}^{8} \text{ " rarr "one of the index laws: } {\left(\frac{a}{b}\right)}^{m} = {\left(\frac{b}{a}\right)}^{-} m$

Now the bases are the same so we have:

$- x = 8 \Leftrightarrow x = - 8$