What is ((4^x+2))/9 <2^{x-1} ?

The solution to ((4^x+2))/9 <2^{x-1} is -1<x<2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((4^x+2))/9 <2^{x-1}

\frac{( 4 ^{x} + 2 )}{9} < 2^{x - 1}

- 1 < x < 2

Interval Notation

(- 1, 2)

Solution steps

\frac{( 4 ^{x} + 2 )}{9} < 2^{x - 1}

Apply exponent rules

\frac{( 2 ^{x} ) ^{2} + 2}{9} < \frac{2 ^{x}}{2}

Let

u = 2^{x}

\frac{u ^{2} + 2}{9} < \frac{u}{2}

\frac{u ^{2} + 2}{9} < \frac{u}{2} : \frac{1}{2} < u < 4

\frac{1}{2} < u < 4

Substitute back

u = 2^{x}

\frac{1}{2} < 2^{x} < 4

a < u < b

then

a < u and u < b

\frac{1}{2} < 2^{x} and 2^{x} < 4

\frac{1}{2} < 2^{x} : x > - 1

2^{x} < 4 : x < 2

Combine the intervals

x > - 1 and x < 2

Merge Overlapping Intervals

- 1 < x < 2

0. 3^{x} < 1 0^{- 4}

2^{n} - 2 \geq 4

x, f > 4^{x} - 2

(0.2)^{\frac{( 2 x - 1 )}{( 1 - x )}} > 5

4^{x} < 2^{x + 1} + 3