What is (2/3)^{-4}m^{-5}n^{-2} ?

The solution to (2/3)^{-4}m^{-5}n^{-2} is (81)/(16m^5n^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (2/3)^{-4}m^{-5}n^{-2}

simplify

(\frac{2}{3})^{- 4} m^{- 5} n^{- 2}

\frac{81}{16 m ^{5} n ^{2}}

Solution steps

(\frac{2}{3})^{- 4} m^{- 5} n^{- 2}

(\frac{2}{3})^{- 4} = \frac{81}{16}

= \frac{81}{16} m^{- 5} n^{- 2}

Apply exponent rule:

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

m^{- 5} = \frac{1}{m ^{5}}

= \frac{81}{16} \cdot \frac{1}{m ^{5}} n^{- 2}

Apply exponent rule:

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

n^{- 2} = \frac{1}{n ^{2}}

= \frac{81}{16} \cdot \frac{1}{m ^{5}} \cdot \frac{1}{n ^{2}}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{81 \cdot 1 \cdot 1}{16 m ^{5} n ^{2}}

Multiply the numbers:

81 \cdot 1 \cdot 1 = 81

= \frac{81}{16 m ^{5} n ^{2}}

5 x + x y + 5 y + y^{2}

4 < \frac{15 x - 8}{15} < 7

∣ 5 x - 15 ∣ - 4 \geq 21

\frac{18}{- 6}

x 3 = 9

What is (2/3)^{-4}m^{-5}n^{-2} ?
The solution to (2/3)^{-4}m^{-5}n^{-2} is (81)/(16m^5n^2)