Question

Solve the following equations.
$lo{g}_{1/5}(2x+5)=lo{g}_{1/5}(16-x^2)+tan\frac{5\pi }{4}$

Answer 1

questuion $lo{g}_{\frac{1}{5}}^{(2x+5)}=lo{g}_{\frac{1}{5}}^{(16-x^2)}+tan\left(\frac{5\pi }{4}\right)$

Solution $tan\left(\frac{5\pi }{4}\right)=1$

We know,

$lo{g}_{\frac{1}{5}}^{(2x+5)}-lo{g}_{\frac{1}{5}}^{(16-x^2)}=1$

$\Rightarrow lo{g}_{\frac{1}{5}}^{\frac{(2x+5)}{(15-x^2)}}=1$

$\Rightarrow \frac{2x+5}{16-x^2}=\frac{1}{5}$

$\Rightarrow x^2+10x+9=0$

$\Rightarrow x^2+x+9x+9=0$

$\Rightarrow (x+1)+9(x+1)=0$

$\Rightarrow (x+1)(x+9)=0$

$x=-1,-9$

$x=-9$ is not valid solution

(log cannot have negative solution)

Answer $:-1$