Question Number 11744 by mnbvcxz89 last updated on 30/Mar/17
![If 5^(n+2) = 625 , find [5(n+3)]^(1/n) .](Q11744.png)
$$\mathrm{If}\:\:\:\mathrm{5}^{\mathrm{n}+\mathrm{2}} =\:\mathrm{625}\:,\:\mathrm{find}\:\left[\mathrm{5}\left(\mathrm{n}+\mathrm{3}\right)\right]^{\mathrm{1}/{n}} . \\ $$
Answered by mrW1 last updated on 30/Mar/17
![5^(n+2) =625=5^4 ⇒n+2=4 ⇒n=2 [5(n+3)]^(1/n) =[5(2+3)]^(1/2) =25^(1/2) =(√(25))=5](Q11746.png)
$$\mathrm{5}^{{n}+\mathrm{2}} =\mathrm{625}=\mathrm{5}^{\mathrm{4}} \\ $$$$\Rightarrow{n}+\mathrm{2}=\mathrm{4} \\ $$$$\Rightarrow{n}=\mathrm{2} \\ $$$$\left[\mathrm{5}\left({n}+\mathrm{3}\right)\right]^{\frac{\mathrm{1}}{{n}}} =\left[\mathrm{5}\left(\mathrm{2}+\mathrm{3}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{25}^{\frac{\mathrm{1}}{\mathrm{2}}} =\sqrt{\mathrm{25}}=\mathrm{5} \\ $$