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

$x = {0.0752499466729842,2.161545847967751}$

How do you solve 5^(x-1)=3x5^(x-1)=3x?