What is (2/3)^{-x^2-2x+12}>= (9/4)^{((x-6))/2} ?

The solution to (2/3)^{-x^2-2x+12}>= (9/4)^{((x-6))/2} is x = 2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2/3)^{-x^2-2x+12}>= (9/4)^{((x-6))/2}

(\frac{2}{3})^{- x^{2} - 2 x + 12} \geq (\frac{9}{4})^{\frac{( x - 6 )}{2}}

x \leq - 3 or x \geq 2

Interval Notation

(- \infty, - 3] \cup [2, \infty)

Solution steps

(\frac{2}{3})^{- x^{2} - 2 x + 12} \geq (\frac{9}{4})^{\frac{( x - 6 )}{2}}

Apply exponent rules

- x^{2} - 2 x + 12 \leq - 2 \cdot \frac{x - 6}{2}

Simplify

- 2 \cdot \frac{x - 6}{2} : - (x - 6)

- x^{2} - 2 x + 12 \leq - (x - 6)

- x^{2} - 2 x + 12 \leq - (x - 6) : x \leq - 3 or x \geq 2

x \leq - 3 or x \geq 2

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

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

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

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

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

What is (2/3)^{-x^2-2x+12}>= (9/4)^{((x-6))/2} ?
The solution to (2/3)^{-x^2-2x+12}>= (9/4)^{((x-6))/2} is x<=-3\lor x>= 2