What is the answer to 2x-4=3^{-x}+1 ?

The answer to 2x-4=3^{-x}+1 is x=(\W_{0}((ln(3))/(2e^{\frac{5ln(3))/2)})}{ln(3)}+5/2

Solve for x: 2x-4=3^{-x}+1

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

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2 x - 4 = 3^{- x} + 1

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

Decimal Notation

x = 2.53100 \dots

Solution steps

2 x - 4 = 3^{- x} + 1

Prepare

2 x - 4 = 3^{- x} + 1

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

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

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

Substitute back

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

solve for

x

Solve

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

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

2^{x} - 1 = 2^{6}

(0.04)^{5 x - x^{2} - 8} = 625

3 = 3^{2 x} - 3^{x} \cdot 2

1 0^{x} = 7000

5 + (3) 4^{x} - 1 = 12

What is the answer to 2x-4=3^{-x}+1 ?
The answer to 2x-4=3^{-x}+1 is x=(\W_{0}((ln(3))/(2e^{\frac{5ln(3))/2)})}{ln(3)}+5/2
Solve for x: 2x-4=3^{-x}+1
The solution is x=(\W_{0}((ln(3))/(2e^{\frac{5ln(3))/2)})}{ln(3)}+5/2