Question

$\frac{\left(125{)}^n{5}^2\right(5^{-\frac{n}{2}}{)}^3-(5^n{)}^{3/2}}{5^{3m}{2}^{3}3}=\frac{1}{25}$
then which of the following is true?

• A. 2 m-n+2=0
• B. n-2m-2=0
• C. 2 m+n-2=0
• D. 3n-6m+4=0

Answer 1

Answer: (D)

3n-6m+4=0

$\frac{{\left(125\right)}^n\left(5^2\right){\left(5^{-\frac{n}{2}}\right)}^3-{\left(5^n\right)}^{3/2}}{{\left(5\right)}^{3m}{\left(2\right)}^3\left(3\right)}=\frac{1}{25}$
$\Rightarrow \frac{{\left(5^3\right)}^n\left(5^2\right)\left(5^{-\frac{3n}{2}}\right)-{\left(5^n\right)}^{3/2}}{{\left(5\right)}^{3m}{\left(2\right)}^3\left(3\right)}=\frac{1}{5^2}$
$\Rightarrow {\left(5^3\right)}^n\left(5^2\right)\left(5^{-\frac{3n}{2}}\right)-{\left(5^n\right)}^{3/2}=\frac{{\left(5\right)}^{3m}{\left(2\right)}^3\left(3\right)}{5^2}$
$\Rightarrow \left(5^{3n}\right)\left(5^2\right)\left(5^{-\frac{3n}{2}}\right)-{\left(5^n\right)}^{3/2}={\left(5\right)}^{-2}{\left(5\right)}^{3m}{\left(2\right)}^3\left(3\right)$
$\Rightarrow \left(5^{{}^{3n-\frac{3n}{2}+2}}\right)-{\left(5^n\right)}^{3/2}={\left(5\right)}^{3m-2}{\left(2\right)}^3\left(3\right)$
$\Rightarrow \left(5^{{}^{\frac{3n+4}{2}}}\right)-{\left(5\right)}^{3n/2}={\left(5\right)}^{3m-2}{\left(2\right)}^3\left(3\right)$
$\Rightarrow \left(5^{{}^{\frac{3n}{2}}}\right)[5^2-1]={\left(5\right)}^{3m-2}{\left(2\right)}^3\left(3\right)$
$\Rightarrow \left(5^{{}^{\frac{3n}{2}}}\right)\left[24\right]={\left(5\right)}^{3m-2}\left(24\right)$
$comparing\exp onentsof5bothsides$
$\frac{3n}{2}=3m-2$
$\Rightarrow 3n=6m-4$
$\Rightarrow 3n-6m+4=0$