What is (125^{(x^2-2)})/(5^x)<= 1/125 ?

The solution to (125^{(x^2-2)})/(5^x)<= 1/125 is (-sqrt(37)+1)/6 <= x<= (sqrt(37)+1)/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(125^{(x^2-2)})/(5^x)<= 1/125

\frac{12 5 ^{(x^{2} - 2)}}{5 ^{x}} \leq \frac{1}{125}

\frac{- 37 + 1}{6} \leq x \leq \frac{37 + 1}{6}

Interval Notation

[\frac{- 37 + 1}{6}, \frac{37 + 1}{6}]

Decimal Notation

- 0.84712 \dots \leq x \leq 1.18046 \dots

Solution steps

\frac{12 5 ^{(x^{2} - 2)}}{5 ^{x}} \leq \frac{1}{125}

Apply exponent rules

\frac{\frac{12 5 ^{x^{2}}}{12 5 ^{2}}}{5 ^{x}} \leq \frac{1}{125}

f (x) > 0

we can multiply or devide both sides of inequality by

f (x)

5^{x}

is greater than

0

for all

x

\frac{\frac{12 5 ^{x^{2}}}{12 5 ^{2}}}{5 ^{x}} \cdot 5^{x} \leq \frac{1}{125} \cdot 5^{x}

Simplify

12 5^{x^{2} - 2} \leq 5^{x - 3}

Apply exponent rules

3 (x^{2} - 2) \leq x - 3

3 (x^{2} - 2) \leq x - 3 : \frac{- 37 + 1}{6} \leq x \leq \frac{37 + 1}{6}

\frac{- 37 + 1}{6} \leq x \leq \frac{37 + 1}{6}

- 5 \leq \frac{4 - 3 x}{2} < 1

\frac{1}{2} \cdot 3

26 n - 4 = 360

x - x^{2} = 0

\frac{7 !}{( 7 - 2 ) ! 2 !}

What is (125^{(x^2-2)})/(5^x)<= 1/125 ?
The solution to (125^{(x^2-2)})/(5^x)<= 1/125 is (-sqrt(37)+1)/6 <= x<= (sqrt(37)+1)/6