Question

Solve the following equations.
$\frac{lo{g}_5(2x+3)}{1-lo{g}_5(2x+3)}+\frac{3}{5-lo{g}_5(2x+3)}=0.$

Answer 1

$\frac{{log}_5(2x+3)}{1-{log}_5(2x+3)}+\frac{3}{5-{log}_5(2x+3)}$

$5{log}_5(2x+3)-5{\left({log}_5(2x+3)\right)}^2+3-3{log}_5(2x+3)=0$

$5{\left({log}_5(2x+3)\right)}^2-2{log}_5(2x+3)+3=0$

Let ${log}_5(2x+3)=y$

${5y}^2-2y+3=0$

${5y}^2-5y-3y+3=0$

$5y(y-1)-3(y-1)=0$

$y=1$ or $5y-3=0$

${log}_5(2x+3)=1$ $y=\frac{3}{5}$

So, $2x+3=5$ or, ${log}_5(2x+3)=\frac{3}{5}$

$2x=2$ $2x+3=5^{3/5}$

$x=1$ $x=\frac{5^{3/5}-3}{2}$