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#