(1/5)^3y=0.008 then $\displaystyle (0.025)^y$ will be
(a)0.25
(b)0.0625
(c)0.125
Given: $\displaystyle (\frac{1}{5})^{3y}=0.008$
Take the log (unspecified base, but same base on both sides.) of both sides: $\displaystyle \log{(\frac{1}{5})^{3y}} = \log{0.008}$
Carry down 3y: $\displaystyle 3y \cdot \log{\frac{1}{5}} = \log{0.008}$
Divide by 3 and $\displaystyle \log{\frac{1}{5}}$ on both sides: $\displaystyle y = \frac{\log{0.008}}{3 \cdot \log{\frac{1}{5}}}$
y = ...
Well, $\displaystyle \tfrac{1}{5}^{3y}=.008\implies \tfrac{1}{125}^y=.008$
However, $\displaystyle .008=\tfrac{1}{125}$, so we get the expression $\displaystyle \tfrac{1}{125}^y=\tfrac{1}{125}$
The only value of y that causes this expression to be true is $\displaystyle y=1$
But then $\displaystyle (0.025)^y\implies(0.025)^1=0.025$...
However, this isn't one of the solutions...
--Chris