What is the answer to 5^{x+1}+5x=6 ?

The answer to 5^{x+1}+5x=6 is x=-(5\W_{0}((e^{\frac{11ln(5))/5}ln(5))/(5))-6ln(5)}{5ln(5)}

Solve for x: 5^{x+1}+5x=6

The solution is x=-(5\W_{0}((e^{\frac{11ln(5))/5}ln(5))/(5))-6ln(5)}{5ln(5)}

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5^{x + 1} + 5 x = 6

x = - \frac{5 W _{0} ( \frac{e ^{\frac{11 l n ( 5 )}{5}} l n ( 5 )}{5} ) - 6 ln ( 5 )}{5 ln ( 5 )}

Decimal Notation

x = 0.07382 \dots

Solution steps

5^{x + 1} + 5 x = 6

Prepare

5^{x + 1} + 5 x = 6

for Lambert form:

1 = (6 - 5 x) e^{l n (5) (- x - 1)}

Rewrite the equation with

(- x + \frac{6}{5}) ln (5) = u

and

x = - \frac{5 u - 6 ln ( 5 )}{5 ln ( 5 )}

1 = (6 - 5 (- \frac{5 u - 6 ln ( 5 )}{5 ln ( 5 )})) e^{l n (5) (- (- \frac{5 u - 6 l n ( 5 )}{5 l n ( 5 )}) - 1)}

Simplify

1 = e^{\frac{5 u - 6 l n ( 5 )}{5} - l n (5)} (6 + \frac{5 u - 6 ln ( 5 )}{ln ( 5 )})

Rewrite

1 = e^{\frac{5 u - 6 l n ( 5 )}{5} - l n (5)} (6 + \frac{5 u - 6 ln ( 5 )}{ln ( 5 )})

in Lambert form:

e^{u} u = \frac{e ^{\frac{11 l n ( 5 )}{5}} ln ( 5 )}{5}

Solve

e^{u} u = \frac{e ^{\frac{11 l n ( 5 )}{5}} ln ( 5 )}{5} : u = W_{0} (\frac{e ^{\frac{11 l n ( 5 )}{5}} ln ( 5 )}{5})

u = W_{0} (\frac{e ^{\frac{11 l n ( 5 )}{5}} ln ( 5 )}{5})

Substitute back

u = (- x + \frac{6}{5}) ln (5),

solve for

x

Solve

(- x + \frac{6}{5}) ln (5) = W_{0} (\frac{e ^{\frac{11 l n ( 5 )}{5}} ln ( 5 )}{5}) : x = - \frac{5 W _{0} ( \frac{e ^{\frac{11 l n ( 5 )}{5}} l n ( 5 )}{5} ) - 6 ln ( 5 )}{5 ln ( 5 )}

x = - \frac{5 W _{0} ( \frac{e ^{\frac{11 l n ( 5 )}{5}} l n ( 5 )}{5} ) - 6 ln ( 5 )}{5 ln ( 5 )}

4 (3^{x + 1}) = 24

4 (3^{x}) = 3 (6^{(3 x - 3)})

71 = 2 7^{x}

4^{- 3 t} = 0.1

4^{(2 x + 1)} = 1024

What is the answer to 5^{x+1}+5x=6 ?
The answer to 5^{x+1}+5x=6 is x=-(5\W_{0}((e^{\frac{11ln(5))/5}ln(5))/(5))-6ln(5)}{5ln(5)}
Solve for x: 5^{x+1}+5x=6
The solution is x=-(5\W_{0}((e^{\frac{11ln(5))/5}ln(5))/(5))-6ln(5)}{5ln(5)}