Question 1075860
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O, great !!!  This is entirely different and fully correct request.


<pre>
After substituting  y = {{{3^x}}}, your equation becomes

y = {{{2 + 1/y}}}

Multiply by "y" both sides. You will get

{{{y^2}}} = {{{2y + 1}}},  or

{{{y^2 - 2y -1}}} = 0.


Apply the quadratic formula to solve the quadratic equation. You will get

{{{y[1,2]}}} = {{{(2 +- sqrt(4 +4))/2}}} = {{{(2+- sqrt(8))/2}}} = {{{1 +- sqrt(2)}}}.


The negative root does not make sense, since y = {{{3^x}}} must be positive. 
It is EXTRANEOUS root for our problem.

The only positive root y = {{{1 + sqrt(2)}}} survives.


So, we have {{{3^x}}} = {{{1+sqrt(2)}}}, which implies

x = {{{log(3, (1+sqrt(2)))}}}.
</pre>

Solved.