What is the answer to 3^x+2x-5=0 ?

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

Solve for x: 3^x+2x-5=0

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

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3^{x} + 2 x - 5 = 0

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

Decimal Notation

x = 1

Solution steps

3^{x} + 2 x - 5 = 0

Prepare

3^{x} + 2 x - 5 = 0

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

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

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

Substitute back

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

solve for

x

Solve

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

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

2^{5 x + 2} = 8^{9}

(\frac{( 2 ^{1 - 2 x} )}{( 2 ^{x + 3} )})^{3} = 4

y, \frac{( 2 ^{y} - 1 )}{( 2 ^{y} + 1 )} = x

3^{x} = 171

2^{1 + x} + 2^{1 - x} = 5

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