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

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Answer 1 · 2015-10-22T17:29:28

$x = \frac{1}{5} (ln 4 - 2)$ $x=1/5(ln4-2)$

$= - 0, 1227411$ $=-0,1227411$

Using laws of exponents and rearranging we may write this as

$e^{5 x} \cdot e^{2} = \frac{8}{2}$ $e^(5x)*e^2=8/2$

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

$5 x = ln (\frac{4}{e^{2}})$ $5x=ln(4/(e^2))$

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

$=-0,1227411$

How do you solve 2e^(5x+2) = 82e5x+2=82e^(5x+2) = 8?