What is the answer to (1+x)sqrt(4)=2^x ?

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

Solve for x: (1+x)sqrt(4)=2^x

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

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(1 + x) 4 = 2^{x}

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

Decimal Notation

x = 3, x = - 0.69009 \dots

Solution steps

(1 + x) 4 = 2^{x}

Prepare

(1 + x) 4 = 2^{x}

for Lambert form:

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

Rewrite the equation with

(- x - 1) ln (2) = u

and

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

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

Simplify

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

Rewrite

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

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 - 1) ln (2),

solve for

x

Solve

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

Solve

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

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

3^{2 x} - 7 \cdot 3^{x} - 18 = 0

2 5^{2 x + 1} = (\frac{1}{9})^{3 x}

2 x + 4^{x} = 6

4^{- 2 x + 2} = 1 6^{x - 3}

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

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