How do you solve #5^(x-1)=3x#?

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Answer 1 · 2017-07-31T20:13:46

See below.

Making #e^lambda = 5# we have

#5^-1 e^(lambda x)= 3x# or

#5^-1/3=x e^(-lambda x) = 1/lambda(lambda x)e^(-lambda x)# then

#(5^-1lambda)/3 = -(-lambdax)e^(-lambda x)# and now using the Lambert function

applying #Y=Xe^X hArr X = W(Y)# we have

#-lambda x = W((-5^-1lambda)/3)# or

#x = -1/lambda W((-5^-1lambda)/3)# with #lambda = log_e 5# or

#x = -(W(-log_e5/15))/log_e5#

The Lambert function furnishes two solutions

#x = {0.0752499466729842,2.161545847967751}#