en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2^n>2n+1,n>= 3

Solution

2^{n} > 2 n + 1, n \geq 3

Solution

No Solution

Solution steps

2^{n} > 2 n + 1

Prepare

2^{n} > 2 n + 1

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

\frac{e ^{\frac{2 u + l n ( 2 )}{2}} u}{e ^{\frac{l n ( 2 )}{2}}} < - \frac{ln ( 2 )}{2 e ^{\frac{l n ( 2 )}{2}}} :

No Solution for

u \in R

No Solution for n \in R

Popular Examples

2^{x+1}>3^{x+2}

2^{x + 1} > 3^{x + 2}

((2^{1-x}-2^x+1))/((2^x-1))=<0

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

(3/4)^{6x+10-x^2}< 27/64

(\frac{3}{4})^{6 x + 10 - x^{2}} < \frac{27}{64}

(5^{x/2})<= (1/5)^{x-3}

(5^{\frac{x}{2}}) \leq (\frac{1}{5})^{x - 3}

1 - x < e^{x}

Frequently Asked Questions (FAQ)

What is 2^n>2n+1,n>= 3 ?
The solution to 2^n>2n+1,n>= 3 is No Solution