Question 176366
Since both of the sides are log base 5, all that is needed is to set the arguments equal to each other:
{{{3x^2-1=2x}}}
{{{3x^2-2x-1}}}
I will use the quadratic formula to solve with a=3,b=-2, c=-1.
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (2 +- sqrt( (-2)^2-4*3*-1 ))/(2*3) }}}
{{{x = (2 +- sqrt(4+12))/6 = (2 +- sqrt(16))/6 = (2 +- 4)/6}}}
So
{{{x=6/6=1}}}
and
{{{x=-2/6=-1/3}}}
Now, with logarithms you always want to check your answer because you cannot take the log of a negative number.  Notice that if you plugged in x=-1/3 to the right hand side (log5(2x)) you would be taking the log of a negative number.  So x=-1/3 is what is called an extraneous solution.  Therefore, x=1 is the only solution.