Question

The solution set of ${\left({\log }_{5}x\right)}^2+{\log }_{5}x+1=\frac{1}{{\log }_{5}x-1}$ is

• A. $(1,3)$
• B. $\varphi$
• C. $\left\{5^{2^{\frac{1}{3}}}\right\}$
• D. $\{1,25\}$

Answer 1

The correct option is C $\left\{5^{2^{\frac{1}{3}}}\right\}$

${\left({\log }_{5}x\right)}^2+{\log }_{5}x+1=\frac{1}{{\log }_{5}x-1}$
Above equation is valid when $x>0,x\ne 1$
Substitute $t={\log }_{5}x$
$t^2+t+1=\frac{1}{t-1}$
$\Rightarrow t^3=2$
$\Rightarrow t=2^{\frac{1}{3}}$
$\Rightarrow {\log }_{5}x=2^{\frac{1}{3}}$
$\Rightarrow x=5^{2^{\frac{1}{3}}}$