Question 657944
<pre>
{{{4^(2y)}}} +{{{ 4^(2y-1)}}} = {{{4}}}

We use the principle that {{{A^(B+C)=A^B*A^C}}} on the second term:

{{{4^(2y)}}} + {{{ 4^(2y)4^(-1)}}} = {{{4}}}

We change the {{{4^(-1)}}} to {{{1/4}}}

{{{4^(2y)}}} +{{{ 4^(2y)*expr(1/4)}}} = {{{4}}}

Multiply through by 4 to remove the fraction:

{{{4*""}}}{{{4^(2y)}}} + {{{4*""}}}{{{ 4^(2y)*expr(1/4)}}} = {{{4*""}}}{{{4}}}

{{{4*""}}}{{{4^(2y)}}} + {{{cross(4)*""}}}{{{ 4^(2y)*expr(1/cross(4))}}} = {{{4*""}}}{{{4}}}

{{{4*""}}}{{{4^(2y)}}} + {{{ 4^(2y)}}} = {{{16}}}

The two terms on the left are like terms and combine as

{{{5*""}}}{{{4^(2y)}}} = {{{16}}}

Then we divide both sides by 5


{{{4^(2y)}}} = {{{16/5}}}

Now we take logs of both sides

{{{log((4^(2y)))}}} = {{{log((16/5))}}}

We use the principle {{{log((A^B)) = B*log((A))}}}

{{{2y*log((4))}}} = {{{log((16/5))}}}

Divide both sides by {{{2*log((4))}}}

{{{y}}} = {{{log((16/5))/(2log((4)))}}} = 0.4195179763

Edwin</pre>