Find the value of y such that 9^y = 3^5/27^3.
Given equation is
\(9y^2={3^5 \over 27^3}\)
⇒\((3y)^2={3^5 \over 3^9}\)
⇒\((3y)^2=3^{-4}\)
⇒\(y^2=3^{-6}\)
⇒\(y^2=(3^{-3})^2\)
⇒\(y=3^{-3}\)
⇒\(y={1 \over 27}\)
∴ Value of y is \({1 \over 27}\).
+526 
