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Find the value of y such that 9^y = 3^5/27^3.

 Apr 26, 2021
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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 \({1 \over 27}\).

 Apr 26, 2021

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