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# help algebra

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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