en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4^x-2*2^x-8<= 0

Solution

4^{x} - 2 \cdot 2^{x} - 8 \leq 0

Solution

x \leq 2

+1

Interval Notation

(- \infty, 2]

Solution steps

4^{x} - 2 \cdot 2^{x} - 8 \leq 0

Apply exponent rules

(2^{x})^{2} - 2 \cdot 2^{x} - 8 \leq 0

Let

u = 2^{x}

u^{2} - 2 u - 8 \leq 0

u^{2} - 2 u - 8 \leq 0 : - 2 \leq u \leq 4

- 2 \leq u \leq 4

Substitute back

u = 2^{x}

- 2 \leq 2^{x} \leq 4

If

a \leq u \leq b

then

a \leq u and u \leq b

- 2 \leq 2^{x} and 2^{x} \leq 4

- 2 \leq 2^{x} :

True for all

x

2^{x} \leq 4 : x \leq 2

Combine the intervals

True for all x and x \leq 2

Merge Overlapping Intervals

x \leq 2

Popular Examples

e^{x} <> 2

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

5^{x + 1} > 3

2. 5^{1 - 2 x} > 1

(7/10)^n<= 20/100

(\frac{7}{10})^{n} \leq \frac{20}{100}

Frequently Asked Questions (FAQ)

What is 4^x-2*2^x-8<= 0 ?
The solution to 4^x-2*2^x-8<= 0 is x<= 2