How do you solve $4^{2 x + 3} = 1$ $4^(2x+3) = 1$ ?

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Answer 1 · 2016-06-25T12:49:31

$x = - \frac{3}{2}$ $x = -3/2$

$4^{2 x + 3} = 1 = 4^{0} \to 2 x + 3 = 0 \to x = - \frac{3}{2}$ $4^(2x+3) = 1= 4^0->2x+3=0->x=-3/2$

How do you solve 4^(2x+3) = 142x+3=14^(2x+3) = 1?