Question

Number of integral value (s) of 'x' satisfying the question
$log{}_{\frac{1}{5}}\left(\right(sec\frac{\pi }{3})x+5)=log{}_5\left(\frac{1}{16-x^2}\right)+tan\frac{13\pi }{4}$ is

• A. $0$
  
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$\text{Solve:}$

${\log }_{\frac{1}{5}}\left(\sec \frac{\pi }{3}\right)x+5)={\log }_5\left(\frac{1}{16-x^2}\right)+\tan \frac{13\pi }{4}$

$\Rightarrow {\log }_5-1(2x+5)={\log }_5\left(\frac{1}{16-x^2}\right)+1$

$[\sec \frac{\pi }{3}=2$

$\tan \left(1\frac{3\pi }{4}\right)=\tan (3\pi +\frac{\pi }{4})=1]$

$\Rightarrow -{\log }_5(2x+5)={\log }_5\left(\frac{1}{16-x^2}\right)+{\log }_{5}5$

$\Rightarrow -{\log }_5(2x+5)={\log }_5\left(\frac{5}{16-x^2}\right)$

$[\log a+\log b=\log ab]$

$\Rightarrow {\log }_5(2x+5)+{\log }_5\left(\frac{5}{16-x^2}\right)=c$

$=>{\log }_5\left(\frac{10x+25}{16-x^2}\right)=0$

$[{\log }_{5}1=0\Rightarrow 5^{\circ }=1]$

$=\frac{10x+25}{16-x^2}=1$

$\Rightarrow 10x+25=10-x^2$

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

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

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

$\Rightarrow x=-1$ or $x=-9$

but, $2x+5>0$ and $\frac{1}{16-x^2}>0$ beccuse

value inside log cannot be negative

or zero

$\Rightarrow 2x+5>0$ and $\frac{1}{16-x^2}>0$

$\Rightarrow x>\frac{-5}{2}\Rightarrow 16-x^2>0=>x^2-16<0$

$\Rightarrow (x-4)(x+4)<0x\in (-4,4)$

So, $x=-1$ is the only one solution

of $x$.