Mathematics - Basic maths Question with Solution | TestHub

Taking log to the base $5$ on both sides

$\left(16{\left({\log }_{5}x\right)}^3-68\left({\log }_{5}x\right)\right)\left({\log }_{5}x\right)=-16$

Let $\left({\log }_{5}x\right)=t$

$16t^4-68t^2+16=0$

$\Rightarrow 4t^4-16t^2-t^2+4=0$

$\Rightarrow \left(4t^2-1\right)\left(t^2-4\right)=0$

$\Rightarrow t=\pm \frac{1}{2},\pm 2$

So ${\log }_{5}x=\pm \frac{1}{2}$ or $\pm 2$

$\Rightarrow x=5^{\frac{1}{2}}, 5^{\frac{-1}{2}}, 5^2, 5^{-2}$

Hence, the product is $\sqrt{5}\times \frac{1}{\sqrt{5}}\times 25\times \frac{1}{25}=1$