Question 1099872
<br>The bases are the same for every term in the equation, so using rules of logarithms we can see
{{{2x+1 = (x-2)/3}}}  
{{{6x+3 = x-2}}}
{{{5x = -5}}}
{{{x = -1}}}<br>
Algebraically, the only possible solution is x = -1; however, that makes the original equation
{{{log(1) = log(-3) - log(3)}}}
and that equation is invalid because log(-3) is undefined.<br>
So there is no value of x that makes the equation true.