Question 131546
<pre><font size = 4 color = "indigo"><b>
Solve for x: 
{{{2*log((x))=log((2))+log((3x-4))}}}

Use this rule of exponents on the left side

                     {{{A*log((B))=log((B^A))}}}

{{{log((x^2))=log((2))+log((3x-4))}}}

Use this rule of logarithms on the right side:

                     {{{log((A)) + log((B)) = log((AB))}}} 

{{{log((x^2))=log((2(3x-4)))}}}

Distribute on the right:

{{{log((x^2))=log((6x-8)))}}}

Use the rule:

               The equation {{{log((A))}}} = {{{log((B))}}} is equivalent to
               the equation {{{A=B}}}

{{{x^2=6x-8}}}

{{{x^2-6x+8=0}}}

{{{(x-2)(x-4)=0}}}

{{{x-2=0}}} gives solution {{{x=2}}}

{{{x-4=0}}} gives solution {{{x=4}}}
                        
Edwin</pre>