Question 136389
So;ve for x:
{{{2log((x+2)) = log((3))+log((2x+1))}}} Apply the "product rule" to the right side. {{{Log[b](M)+Log[b](N) = Log[b](MN)}}}
{{{2log((x+2)) = log((3(2x+1)))}}} Apply the "power rule" to the left side.{{{pLog[b](M) = Log[b](M)^p}}}
{{{log((x+2))^2 = log((3(2x+1)))}}} therefore:
{{{(x+2)^2 = 3(2x+1)}}} Simplify.
{{{x^2+4x+4 = 6x+3}}} Subtract (6x+3) from both sides.
{{{x^2-2x+1 = 0}}} Factor.
{{{(x-1)(x-1) = 0}}}, so the solution is a double root:
{{{x = 1}}}
Check:
{{{2log((x+2)) = log((3))+log((2x+1))}}} Substitute x = 1.
{{{2log((1+2)) = log((3))+log((2(1)+1))}}} Simplify.
{{{2log((3)) = log((3))+log((3))}}}
{{{2log((3)) = 2log((3))}}}