Question

If $5^{k^2}\left({25}^{2k}\right)\left(625\right)=25\sqrt{5}$ and $k<-1$, find the value of $k$.

• A. $-3.581$
• B. $-3.162$
• C. $-2.613$
• D. $-1.581$
• E. $-0.419$

Answer 1

The correct option is A $-3.581$

Given, $5^{k^2}\left({25}^{2k}\right)\left(625\right)=25\sqrt{5}$
$5^{k^2+4k+4}=5^{\frac{5}{2}}$

By comparing powers, we get
$k^2+4k+4=\frac{5}{2}$ which implies ${2k}^2+8k+3=0$

$\Rightarrow k=-3.581$ ...(given $k<-1$)