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

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

Solve for x: (2^{x+1})/(4^{x-1)}=2x

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{W _{0} ( 4 ln ( 2 ) )}{ln ( 2 )}

Decimal Notation

x = 1.45699 \dots

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

x ln (2) = u

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

e^{u} u = 4 ln (2)

Solve

e^{u} u = 4 ln (2) : u = W_{0} (4 ln (2))

u = W_{0} (4 ln (2))

Substitute back

u = x ln (2),

solve for

x

Solve

x ln (2) = W_{0} (4 ln (2)) : x = \frac{W _{0} ( 4 ln ( 2 ) )}{ln ( 2 )}

x = \frac{W _{0} ( 4 ln ( 2 ) )}{ln ( 2 )}

Verify Solutions:

x = \frac{W _{0} ( 4 ln ( 2 ) )}{ln ( 2 )}

True

The solution is

x = \frac{W _{0} ( 4 ln ( 2 ) )}{ln ( 2 )}

Verify Solutions:

x = \frac{W _{0} ( 4 ln ( 2 ) )}{ln ( 2 )}

True

The solution is

x = \frac{W _{0} ( 4 ln ( 2 ) )}{ln ( 2 )}

e^{2 x} - x^{3} = 0

(\frac{1}{2})^{(x - 2)} + 1 = 0

3 6^{x + 1} = 6^{x^{2} + 6 x - 10}

- 4 = e^{x}

e^{2 x} + 9 e^{- 2 x} - 10 = 0

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