How do you solve $2e^(5x+2) = 8$ ?

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Answer 1 · 2015-10-22T17:29:28

$x=1/5(ln4-2)$

$=-0,1227411$

Using laws of exponents and rearranging we may write this as

$e^(5x)*e^2=8/2$

Now taking the natural logarithm on both sides and using laws of logs we get

$5x=ln(4/(e^2))$

$thereforex=1/5(ln4-2)$

$=-0,1227411$

How do you solve 2e^(5x+2) = 82e^(5x+2) = 8?