$
\eqalign{
& 4\log (2x + 2) = - 2 \cr
& \log (2x + 2) = - \frac{1}
{2} \cr
& 2x + 2 = 10^{ - \frac{1}
{2}} \cr
& 2x + 2 = \frac{1}
{{10^{\frac{1}
{2}} }} \cr
& 2x + 2 = \frac{1}
{{\sqrt {10} }} \cr
& 2x = \frac{1}
{{\sqrt {10} }} - 2 \cr
& x = \frac{1}
{{2\sqrt {10} }} - 1 \cr
& x = \frac{{\sqrt {10} }}
{{20}} - 1 \cr}
$ |
$
\eqalign{
& ^4 \log (2x + 2) = - 2 \cr
& 2x + 2 = 4^{ - 2} \cr
& 2x + 2 = \frac{1}
{{4^2 }} \cr
& 2x + 2 = \frac{1}
{{16}} \cr
& 2x = - 1\frac{{15}}
{{16}} \cr
& x = - \frac{{31}}
{{32}} \cr}
$ |
