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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Decimal Notation

x = - 2.53100 \dots

Solution steps

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

Prepare

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

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{2 u + 5 ln ( 3 )}{2 ln ( 3 )}

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

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:

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

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

16 (4^{2 x}) + 63 (4^{x}) - 4 = 0

5^{x + 2} + 5^{2 x + 2} - 50 = 0

(- 9)^{(1 - 2 x)} = 3^{- 4}

8^{x + 3} = 3 2^{3 x}

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