Find the value of y given that
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18 Apr 2021, 21:53
To solve for y, we need to equate powers.
\(\frac{4^{y^2} }{ 64} = 2^ {-y}\)
\(\frac{2^{2y^2} }{ 2^6 * 2^{-y}} = 1\) (Any number to the power 0 is 1)
\(2^{2y^2 - 6 + y} = 2^0\)
Equating powers,
\(2y^2 - 6 + y = 0\)
We will have to factorise this expression.
\(2y^2 + 4y - 3y - 6 = 0\)
\(2y (y + 2) - 3(y + 2) = 0\)
\((2y - 3) (y + 2) = 0\)
y = 3/2, -2
Since y<|y|, only -2 can satisfy this.
Hence, -2 is the answer.