Question

Sometimes to solve an equation, we may use the identity $a^{{\log }_{a}b}=b,b>0,a>0,a\ne 1$

Then the number of solution(s) of $x^{{\log }_x{(x+3)}^2}=16$ is/are,

• A. $0$
• B. $1$
• C. $2$
• D. infinite
• E. None of these

Answer 1

The correct option is A $0$

$x^{{\log }_x{(x+3)}^2}=16$

Above equation is valid when $x>0,x\ne 1$
$x^{{\log }_x{(x+3)}^2}=16$

$\Rightarrow (x+3{)}^2=16=4^2,[\because a^{{\log }_{a}b}=b]$

$\Rightarrow x+3=\pm 4$

$\Rightarrow x=1,-7$

Since, $x>0,x\ne 1$

Therefore, $x=\left\{\varnothing \right\}$