Question

If $a,b>0,a,b\ne 1,c>0$, then ${\log }_{a}c=\frac{{\log }_{b}c}{{\log }_{b}a}=\left({\log }_{b}c\right)\left({\log }_{a}b\right)$

The number of solution(s) of the equation ${\log }_5(x^4+5)-{\log }_{1/5}(x^2+25)=\frac{3}{2}$ is/are

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

Answer 1

The correct option is A $0$

Given that, ${\log }_5(x^4+5)-{\log }_{\frac{1}{5}}(x^2+25)=\frac{3}{2}$ ....(1)
$\Rightarrow {\log }_5(x^4+5)+{\log }_5(x^2+25)=\frac{3}{2}[\because {\log }_{1/a}b=-{\log }_{a}b]$
$\Rightarrow {\log }_5(x^4+5)(x^2+25)=\frac{3}{2}[\because \log a+\log b=\log (ab\left)\right]$
Clearly, $x^4,x^2\ge 0$
Therefore, $(x^4+5)(x^2+25)\ge 125$
$\Rightarrow {\log }_5(x^4+5)(x^2+25)\ge 3$
$\Rightarrow \frac{3}{2}\ge 3$
This is an ambiguous equation.
Therefore equation (1) has zero solution.