What is (3^{x^2-1}-27^{x+1})(log_{3}(x+8)-2)<= 0 ?

The solution to (3^{x^2-1}-27^{x+1})(log_{3}(x+8)-2)<= 0 is -8<x<=-1\lor 1<= x<= 4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(3^{x^2-1}-27^{x+1})(log_{3}(x+8)-2)<= 0

(3^{x^{2} - 1} - 2 7^{x + 1}) (lo g_{3} (x + 8) - 2) \leq 0

- 8 < x \leq - 1 or 1 \leq x \leq 4

Interval Notation

(- 8, - 1] \cup [1, 4]

Solution steps

(3^{x^{2} - 1} - 2 7^{x + 1}) (lo g_{3} (x + 8) - 2) \leq 0

Factor

(3^{x^{2} - 1} - 2 7^{x + 1}) (lo g_{3} (x + 8) - 2) : (3^{(x + 1) (x - 1)} - 2 7^{x + 1}) (lo g_{3} (x + 8) - 2)

(3^{(x + 1) (x - 1)} - 2 7^{x + 1}) (lo g_{3} (x + 8) - 2) \leq 0

Identify the intervals

- 8 < x \leq - 1 or 1 \leq x \leq 4

lo g_{3} (x^{2}) > 2

ln (x) < 1

lo g_{10} (x) (\frac{( 2 x + \frac{2}{5} )}{( 5 ( 1 - x ) )}) > 0

lo g_{4} (x^{2} - 3 x + 2) > lo g_{4} (x^{2} - 5 x + 4)

lo g_{2} (2 + x) < 1

What is (3^{x^2-1}-27^{x+1})(log_{3}(x+8)-2)<= 0 ?
The solution to (3^{x^2-1}-27^{x+1})(log_{3}(x+8)-2)<= 0 is -8<x<=-1\lor 1<= x<= 4