How do you solve #3^x= 3x+5#?

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Answer 1 · 2018-04-03T15:34:07

See below.

Making #y = 3x+5# we have

#3^((y-5)/3) = y# or

#3^(y/3)3^(-5/3) = y#

now making #3^(1/3) = e^lambda# we have

#e^(lambda y) 3^(-5/3) = y# or

#3^(-5/3)=y e^(-lambda y)# or

#(-lambda)3^(-5/3) = (-lambda y) e^(-lambda y)#

At this point we use the Lambert function

#Y = X e^X hArr X = W(Y)# and then

#-lambda y = W((-lambda)3^(-5/3))# or

#y = -1/lambda W((-lambda)3^(-5/3)) = 3x-5# and finally

#x = (-5 Log 3 - 3 W(-(Log 3/(9 xx 3^(2/3)))))/(3 Log 3)#

We have then two solutions

#x =-1.60981# and #x = 2.24048#

Attached a plot showing the intersections of #y = 3^x# and #y = 3x+5#

enter image source here