How do you solve $5^{x - 1} = 3 x$ $5^(x-1)=3x$ ?

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Answer 1 · 2017-07-31T20:13:46

See below.

Making $e^{λ} = 5$ $e^lambda = 5$ we have

$5^{- 1} e^{λ x} = 3 x$ $5^-1 e^(lambda x)= 3x$ or

$\frac{5^{- 1}}{3} = x e^{- λ x} = \frac{1}{λ} (λ x) e^{- λ x}$ $5^-1/3=x e^(-lambda x) = 1/lambda(lambda x)e^(-lambda x)$ then

$\frac{5^{- 1} λ}{3} = - (- λ x) e^{- λ x}$ $(5^-1lambda)/3 = -(-lambdax)e^(-lambda x)$ and now using the Lambert function

applying $Y = X e^{X} \Leftrightarrow X = W (Y)$ $Y=Xe^X hArr X = W(Y)$ we have

$- λ x = W (\frac{- 5^{- 1} λ}{3})$ $-lambda x = W((-5^-1lambda)/3)$ or

$x = - \frac{1}{λ} W (\frac{- 5^{- 1} λ}{3})$ $x = -1/lambda W((-5^-1lambda)/3)$ with $λ = {log}_{e} 5$ $lambda = log_e 5$ or

$x = - \frac{W (- \frac{{log}_{e} 5}{15})}{{log}_{e} 5}$ $x = -(W(-log_e5/15))/log_e5$

The Lambert function furnishes two solutions

$x = {0.0752499466729842, 2.161545847967751}$ $x = {0.0752499466729842,2.161545847967751}$

How do you solve 5x−1=3x5^(x-1)=3x?