What is the answer to 5^{(-x)}+7=2x+4 ?

The answer to 5^{(-x)}+7=2x+4 is x=(\W_{0}((ln(5))/(2e^{\frac{3ln(5))/2)})}{ln(5)}+3/2

Solve for x: 5^{(-x)}+7=2x+4

The solution is x=(\W_{0}((ln(5))/(2e^{\frac{3ln(5))/2)})}{ln(5)}+3/2

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5^{(- x)} + 7 = 2 x + 4

x = \frac{W _{0} ( \frac{l n ( 5 )}{2 e ^{\frac{3 l n ( 5 )}{2}}} )}{ln ( 5 )} + \frac{3}{2}

Solution steps

5^{(- x)} + 7 = 2 x + 4

Prepare

5^{(- x)} + 7 = 2 x + 4

for Lambert form:

1 = (2 x - 3) e^{l n (5) x}

Rewrite the equation with

(x - \frac{3}{2}) ln (5) = u

and

x = \frac{u}{ln ( 5 )} + \frac{3}{2}

1 = (2 (\frac{u}{ln ( 5 )} + \frac{3}{2}) - 3) e^{l n (5) (\frac{u}{l n ( 5 )} + \frac{3}{2})}

Simplify

1 = \frac{2 e ^{u + \frac{3}{2} l n (5)} u}{ln ( 5 )}

Rewrite

1 = \frac{2 e ^{u + \frac{3}{2} l n (5)} u}{ln ( 5 )}

in Lambert form:

\frac{e ^{\frac{2 u + 3 l n ( 5 )}{2}} u}{e ^{\frac{3 l n ( 5 )}{2}}} = \frac{ln ( 5 )}{2 e ^{\frac{3 l n ( 5 )}{2}}}

Solve

\frac{e ^{\frac{2 u + 3 l n ( 5 )}{2}} u}{e ^{\frac{3 l n ( 5 )}{2}}} = \frac{ln ( 5 )}{2 e ^{\frac{3 l n ( 5 )}{2}}} : u = W_{0} (\frac{ln ( 5 )}{2 e ^{\frac{3 l n ( 5 )}{2}}})

u = W_{0} (\frac{ln ( 5 )}{2 e ^{\frac{3 l n ( 5 )}{2}}})

Substitute back

u = (x - \frac{3}{2}) ln (5),

solve for

x

Solve

(x - \frac{3}{2}) ln (5) = W_{0} (\frac{ln ( 5 )}{2 e ^{\frac{3 l n ( 5 )}{2}}}) : x = \frac{W _{0} ( \frac{l n ( 5 )}{2 e ^{\frac{3 l n ( 5 )}{2}}} )}{ln ( 5 )} + \frac{3}{2}

x = \frac{W _{0} ( \frac{l n ( 5 )}{2 e ^{\frac{3 l n ( 5 )}{2}}} )}{ln ( 5 )} + \frac{3}{2}

8^{- 5 a} - 5 = 53

y, x = y e^{y}

m = - 3 (5)^{n + 4} - \frac{1}{2}

3^{x} = 45

8^{y} = 16

What is the answer to 5^{(-x)}+7=2x+4 ?
The answer to 5^{(-x)}+7=2x+4 is x=(\W_{0}((ln(5))/(2e^{\frac{3ln(5))/2)})}{ln(5)}+3/2
Solve for x: 5^{(-x)}+7=2x+4
The solution is x=(\W_{0}((ln(5))/(2e^{\frac{3ln(5))/2)})}{ln(5)}+3/2