How do you solve $4^{2 - 5 x} = \frac{1}{64}$ $4^(2-5x)=1/64$ ?

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How do you solve $4^{2 - 5 x} = \frac{1}{64}$ $4^(2-5x)=1/64$ ?

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Answer 1 · 2016-05-29T05:50:00

Notice $\frac{1}{64} = 4^{- 3}$ $1/64=4^-3$ hence

$4^{2 - 5 x} = 4^{- 3} \Rightarrow log (4^{2 - 5 x}) = {log 4}^{- 3} \Rightarrow (2 - 5 x) = - 3 \Rightarrow 5 = 5 x \Rightarrow x = 1$ $4^(2-5x)=4^-3=>log(4^(2-5x))=log4^-3=>(2-5x)=-3=>5=5x=>x=1$

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How do you solve $4^{2 - 5 x} = \frac{1}{64}$ $4^(2-5x)=1/64$ ?

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