How do you solve $2^{x + 2} - 2^{x + 5} = - 7$ $2^(x+2) - 2^(x+5)= -7$ ?

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How do you solve $2^{x + 2} - 2^{x + 5} = - 7$ $2^(x+2) - 2^(x+5)= -7$ ?

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Answer 1 · 2015-09-09T16:46:42

We have that $2^{x + 2} - 2^{x + 5} = - 7 \Rightarrow 2^{x} (2^{2} - 2^{5}) = - 7 \Rightarrow 2^{x} \cdot (- 28) = - 7 \Rightarrow 2^{x} = \frac{7}{28} \Rightarrow 2^{x} = \frac{1}{4} \Rightarrow 2^{x} = 2^{- 2} \Rightarrow x = - 2$ $2^(x+2)-2^(x+5)=-7=>2^x(2^2-2^5)=-7=>2^x*(-28)=-7=>2^x=7/28=>2^x=1/4=>2^x=2^(-2)=>x=-2$

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How do you solve $2^{x + 2} - 2^{x + 5} = - 7$ $2^(x+2) - 2^(x+5)= -7$ ?

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How do you solve $2^{x + 2} - 2^{x + 5} = - 7$ $2^(x+2) - 2^(x+5)= -7$ ?