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

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

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

The solution is x=1,x=-(\W_{0}(-(ln(2))/4)+3ln(2))/(ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = 1, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{4} ) + 3 ln ( 2 )}{ln ( 2 )}

Decimal Notation

x = 1, x = - 2.69009 \dots

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

(- x - 3) ln (2) = u

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

e^{u} u = - \frac{ln ( 2 )}{4}

Solve

e^{u} u = - \frac{ln ( 2 )}{4} : u = - 4 ln (2), u = W_{0} (- \frac{ln ( 2 )}{4})

u = - 4 ln (2), u = W_{0} (- \frac{ln ( 2 )}{4})

Substitute back

u = (- x - 3) ln (2),

solve for

x

Solve

(- x - 3) ln (2) = - 4 ln (2) : x = 1

Solve

(- x - 3) ln (2) = W_{0} (- \frac{ln ( 2 )}{4}) : x = - \frac{W _{0} ( - \frac{l n ( 2 )}{4} ) + 3 ln ( 2 )}{ln ( 2 )}

x = 1, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{4} ) + 3 ln ( 2 )}{ln ( 2 )}

4 (2^{x} + 2^{- x}) = 17

2^{2 x} + 2^{x} - 42 = 0

6^{x + 1} = 216

e^{\frac{5 y - 3}{4}} = e^{\frac{4 + y}{3}}

0. 5^{x + 7} \times 0. 5^{1 - 2 x} = 2

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