Question 1134204
<pre>
{{{6^(3x-4)*2^(2x+3)=648(48)^x}}}

Take logs of both sides  (can take natural logs instead if preferred)

{{{log((6^(3x-4)*2^(2x+3)))=log((648(48)^x))}}}

Use the product-to-sum log fact on both sides:

{{{log((6^(3x-4)))+log((2^(2x+3)))=log((648^""))+log((48^x)))}}}

Use the exponent-to-multiplier log fact on both sides:

{{{(3x-4)log((6^""))+(2x+3)log((2^""))=log((648^""))+x*log((48^"")))}}}

Let's break three of those logs down to the side:

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

     2. {{{log((648^""))=log((2^3*3^4))=log((2^3))+log((3^4))=3*log((2^""))+4*log((3^""))}}}

     3. {{{log((48^""))=log((2^4*3))=log((2^4))+log((3^""))=4*log((2^""))+log((3^""))}}} 

Those are all in terms of log(2) and Log(3). 

To keep from getting bogged down with too many complicated symbols,

let A = log(2) and let B = log(3), then

     1.  {{{log((6^""))=log((2^""))+log((3^""))=A+B}}}

     2. {{{log((648^""))=3*log((2^""))+4*log((3^""))=3A+4B}}}

     3. {{{log((48^""))=4*log((2^""))+log((3^""))=4A+B}}} 


Now go back to the equation:

{{{(3x-4)log((6^""))+(2x+3)log((2^""))=log((648^""))+x*log((48^"")))}}}

and substitute the expressions in A and B for the logs:

{{{(3x-4)(A+B)+(2x+3)(A)=(3A+4B)+x*(4A+B)}}}

Multiply out:

{{{(3Ax+3Bx-4A-4B)+(2Ax+3A)=3A+4B+4Ax+Bx}}}

{{{3Ax+3Bx-4A-4B+2Ax+3A=3A+4B+4Ax+Bx}}}

Combine like terms on the left:

{{{5Ax+3Bx-A-4B=3A+4B+4Ax+Bx}}}

Get all the terms in x on the left and all the others on the right:

{{{5Ax+3Bx-4Ax-Bx=A+4B+3A+4B}}}

Combine like terms:

{{{Ax+2Bx=4A+8B}}}

Factor out x on the left side and 4 out of the right side:

{{{x(A+2B)=4(A+2B)}}}

Divide both sides by (A+2B)

{{{x=4}}}

Edwin</pre>