How do you solve $4^{x + 1} = \frac{1}{64}$ $4^(x+1)=1/64$ ?

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Answer 1 · 2016-02-08T23:14:44

$x = - 4$ $x=-4$

$4^{- 3} = \frac{1}{4^{3}} = \frac{1}{64}$ $4^(-3) = 1/(4^3) = 1/64$ . So, $x + 1 = - 3, or, x = - 4$ $x+1 = -3, or, x=-4$ .

How do you solve 4x+1=1644^(x+1)=1/64?