What is (t^2)^{-2}(t^2)^{-5} ?

The solution to (t^2)^{-2}(t^2)^{-5} is 1/(t^{14)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (t^2)^{-2}(t^2)^{-5}

simplify

(t^{2})^{- 2} (t^{2})^{- 5}

\frac{1}{t ^{14}}

Solution steps

(t^{2})^{- 2} (t^{2})^{- 5}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

(t^{2})^{- 2} (t^{2})^{- 5} = (t^{2})^{- 2 - 5}

= (t^{2})^{- 2 - 5}

Subtract the numbers:

- 2 - 5 = - 7

= (t^{2})^{- 7}

Apply exponent rule:

(a^{b})^{c} = a^{b c},

assuming

a \geq 0

= t^{2 (- 7)}

2 (- 7) = - 14

= t^{- 14}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

t^{- 14} = \frac{1}{t ^{14}}

= \frac{1}{t ^{14}}

9 x - 4 = 1

\frac{( 16 x ^{4} y ^{6} ) ^{\frac{1}{2}}}{x ^{5} y}

(x - 1) (x + 4) (x - 5) \geq 0

4 - 4 x^{2} = 0

\frac{x}{3} \geq 2 + \frac{x}{6}

What is (t^2)^{-2}(t^2)^{-5} ?
The solution to (t^2)^{-2}(t^2)^{-5} is 1/(t^{14)}