How do you solve $3^{x} = 3 x + 5$ $3^x= 3x+5$ ?

Question

1789 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-03T15:34:07

See below.

Making $y = 3 x + 5$ $y = 3x+5$ we have

$3^{\frac{y - 5}{3}} = y$ $3^((y-5)/3) = y$ or

$3^{\frac{y}{3}} 3^{- \frac{5}{3}} = y$ $3^(y/3)3^(-5/3) = y$

now making $3^{\frac{1}{3}} = e^{λ}$ $3^(1/3) = e^lambda$ we have

$e^{λ y} 3^{- \frac{5}{3}} = y$ $e^(lambda y) 3^(-5/3) = y$ or

$3^{- \frac{5}{3}} = y e^{- λ y}$ $3^(-5/3)=y e^(-lambda y)$ or

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

At this point we use the Lambert function

$Y = X e^{X} \Leftrightarrow X = W (Y)$ $Y = X e^X hArr X = W(Y)$ and then

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

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

$x = \frac{- 5 log 3 - 3 W (- (\frac{log 3}{9 \times 3^{\frac{2}{3}}}))}{3 log 3}$ $x = (-5 Log 3 - 3 W(-(Log 3/(9 xx 3^(2/3)))))/(3 Log 3)$

We have then two solutions

$x = - 1.60981$ $x =-1.60981$ and $x = 2.24048$ $x = 2.24048$

Attached a plot showing the intersections of $y = 3^{x}$ $y = 3^x$ and $y = 3 x + 5$ $y = 3x+5$

enter image source here

How do you solve 3^x= 3x+53x=3x+53^x= 3x+5?