Question 709009
I'll take this part of the way through the solution steps.


You seem to ask to solve, {{{2^x=6*3^x}}}


I'll use "natural logs" because it's less effort for the typesetting on this algebra.com system, less missing grouping symbols in the text needed.


{{{ln(2^x)=ln(6)+ln(3^x)}}}
{{{xln(2)=ln(6)+xln(3)}}}
{{{-ln(6)=xln(3)-xln(2)}}}
{{{-ln(6)=x(ln(3)-ln(2))}}}
{{{(-ln(6))/(ln(3)-ln(2))=x}}}, and I stop here, so maybe you could finish.  This is an expression using only constants equal to the variable, x.