How do you solve $3^{- 4 x + 4} = 9$ $3 ^(-4x+4) = 9$ ?

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Answer 1 · 2016-05-23T07:55:41

$x = \frac{1}{2}$ $x=1/2$

$3^{- 4 x + 4} = 9$ $3^(-4x+4)=9$

$3^{- 4 x} \cdot 3^{4} = 3^{2}$ $3^(-4x) *3^4=3^2$

$3^{- 4 x} = \frac{3^{2}}{3^{4}}$ $3^(-4x)=(3^2)/(3^4)$

$3^{- 4 x} = 3^{2} \cdot 3^{- 4}$ $3^(-4x)=3^2*3^-4$

$3^{- 4 x} = 3^{- 2}$ $color(red)(3)^(-4x)=color(red)(3)^-2$

$Both side of equation has the same base value$ $"Both side of equation has the same base value"$

$- 4 x = - 2$ $-4x=-2$

$x = \frac{2}{4}$ $x=2/4$

$x = \frac{1}{2}$ $x=1/2$

How do you solve 3−4x+4=93 ^(-4x+4) = 9?