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 ${\left({\log }_{x}5\right)}^3-{\left({\log }_{x}5\right)}^2-3{\log }_{\sqrt{x}}5=0$ is/are

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

Answer 1

Answer: (B)

$2$

${\left({\log }_{x}5\right)}^3-{\left({\log }_{x}5\right)}^2-3{\log }_{\sqrt{x}}5=0$
${\left({\log }_{x}5\right)}^3-{\left({\log }_{x}5\right)}^2-6{\log }_{x}5=0[\because {\log }_{a^m}b=\frac{1}{m}{\log }_{a}b]$
Above equation is valid when$x>0,x\ne 1$.
Put ${\log }_{x}5=t$, to obtain
$t^3-t^2-6t=0$
$\Rightarrow t(t^2-t-6)=0$
$\Rightarrow t(t-3)(t+2)=0\Rightarrow t=0,3,-2$
$\Rightarrow 5=x^0,x^3,x^{-2}$
But $5=x^0$ is not possible
Therefore, $x=\{\sqrt[3]{35},\frac{1}{\sqrt{5}}\}$