How do you solve $5(2^y)=3-(2^(y+2))$ ?

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Answer 1 · 2016-05-29T14:33:51

$y =-log_2 3$

$5times 2^y-3+4times2^y=0->9times 2^y-3=0$ .

Applying logarithm to both sides

$y log_e 2=log_e(1/3)$ .

Solving for $y$

$y = (log_e(1/3))/(log_e 2) = -log_2 3$

How do you solve 5(2^y)=3-(2^(y+2))5(2^y)=3-(2^(y+2))?