Question 1031793
<pre><b>
{{{3^(3x+1)}}}{{{""=""}}}{{{6^(x+2)}}}

{{{log((3^((3x+1))))}}}{{{""=""}}}{{{log((6^((x+2))))}}}

{{{(3x+1)log((3))}}}{{{""=""}}}{{{(x+2)log((6))}}}

Let A = log(3) and let B = log(6)

{{{(3x+1)A}}}{{{""=""}}}{{{(x+2)B}}}

{{{A(3x+1)}}}{{{""=""}}}{{{B(x+2)}}}

{{{3Ax+A}}}{{{""=""}}}{{{Bx+2B}}}

{{{3Ax-Bx}}}{{{""=""}}}{{{2B-A}}}

{{{x(3A-B)}}}{{{""=""}}}{{{2B-A}}}

{{{x(3A-B)/(3A-B)}}}{{{""=""}}}{{{(2B-A)/(3A-B)}}}

{{{x(cross(3A-B))/(cross(3A-B))}}}{{{""=""}}}{{{(2B-A)/(3A-B)}}}

{{{x}}}{{{""=""}}}{{{(2B-A)/(3A-B)}}}

Substitute log(3) for A and log(6) for B

{{{x}}}{{{""=""}}}{{{(2log((6))-log((3)))/(3log((3))-log((6)))}}}

Get calculator:
 
{{{x}}}{{{""=""}}}{{{(2(0.7781512504)-(0.4771212547))/(3(0.4771212547)-(0.7781512504))}}}

{{{x}}}{{{""=""}}}{{{1.652113552}}}

Edwin</pre></b>