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

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

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Answer 1 · 2015-09-09T21:24:15

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

Explanation:

Well it is $e^{4 x - 1} = e^{2 x} \Rightarrow e^{4 x - 2 x - 1} = 1 \Rightarrow e^{2 x - 1} = 1 \Rightarrow 2 x - 1 = 0 \Rightarrow x = \frac{1}{2}$ $e^(4x-1)=e^(2x)=>e^(4x-2x-1)=1=>e^(2x-1)=1=>2x-1=0=>x=1/2$

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

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How do you solve e^(4x-1)=(e^2)^x e4x−1=(e2)xe^(4x-1)=(e^2)^x ?

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