Question

Solve $1+2{\log }_{x+2}5={\log }_5(x+2)$

Answer 1

$1+2{\log }_{x+2}5={\log }_5\left(x+2\right)$

$=1+\frac{2}{{\log }_{5}x+2}={\log }_5\left(x+2\right)$

$Let{\log }_{5}x+2=y$

$1+\frac{2}{y}=y$

$y+2=y^2$

$y^2-y-2=0$

$\left(y+1\right)\left(y-2\right)=0$

$y=-1,,y=2$

$ify=-1$

${\log }_{5}x+2=-1$

$x+2=5^{{}^{-1}}$

$x+2=\frac{1}{5}-2$

$x=\frac{-9}{5}$

$ify=2$

${\log }_5\left(x+2\right)=2$

$x+2=5^2$

$x=25-2$

$x=23$

$sox=\frac{-9}{5},23$