Question 890920
{{{2*ln(y+1) = ln(5(y^2 - 1))}}}
{{{ln((y+1)^2) = ln(5(y^2 - 1))}}}
{{{(y+1)^2=5(y^2-1)}}}
{{{y^2+2y+1=5y^2-5}}}
{{{4y^2-2y-6=0}}}
{{{2y^2-y-3=0}}}
{{{(y+1)(2y-3)=0}}}
Two possible solutions:
{{{y+1=0}}}
{{{y=-1}}}
Not allowed since {{{ln(0)}}} is not defined.
{{{2y-3=0}}}
{{{2y=3}}}
{{{highlight(y=3/2)}}}