Question 20730
{{{1-logx = log((3x-1))}}} Add logx to both sides of the equation.
{{{1 = log((3x-1)) + logx}}} Apply the product rule for logarithms.
{{{1 = log(((3x-1)*(x)))}}}
{{{1 = log((3x^2-x))}}} Recall that: If {{{log(a) = 1}}}, then {{{a = 10}}} because: {{{log(10)=1}}}, therefore:
{{{(3x^2-x) = 10}}} Subtract 10 from both sides.
{{{3x^2-x-10= 0}}} Solve this quadratic equation by factoring.
{{{(3x+5)(x-2) = 0}}} Apply the zero products principle.
{{{(3x+5) = 0}}} and/or {{{(x-2) = 0}}}
If {{{3x+5 = 0}}} then {{{3x = -5}}} and {{{x = -5/3}}}
If {{{x-2 = 0}}} then {{{x = 2}}}

The roots are: 
{{{x = 2}}}
{{{x = (-5/3)}}}

Check:
1) x = 2
{{{1-log(2) = log((3(2)-1))}}}
{{{1-0.30103 = log(5)}}}
{{{0.69897 = 0.69897}}}

2) {{{x = -5/3}}}
1-log((-5/3)) = 0.77815..., -1.36437...
log(3(-5/3)-1) = 0.77815..., 1.36437...  
{{{x = (-5/3)}}} is not a valid root.

Answer is x = 2