What is (a^2b^3)^{-2}(a^5b^4)^{-3} ?

The solution to (a^2b^3)^{-2}(a^5b^4)^{-3} is 1/(a^{19)b^{18}}

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simplify (a^2b^3)^{-2}(a^5b^4)^{-3}

simplify

(a^{2} b^{3})^{- 2} (a^{5} b^{4})^{- 3}

\frac{1}{a ^{19} b ^{18}}

Solution steps

(a^{2} b^{3})^{- 2} (a^{5} b^{4})^{- 3}

Simplify

(a^{2} b^{3})^{- 2} : \frac{1}{a ^{4} b ^{6}}

= \frac{1}{a ^{4} b ^{6}} (a^{5} b^{4})^{- 3}

Simplify

(a^{5} b^{4})^{- 3} : \frac{1}{a ^{15} b ^{12}}

= \frac{1}{a ^{4} b ^{6}} \cdot \frac{1}{a ^{15} b ^{12}}

Apply the fraction rule:

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

= \frac{1 \cdot 1}{a ^{4} b ^{6} a ^{15} b ^{12}}

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{a ^{4} b ^{6} a ^{15} b ^{12}}

Simplify

a^{4} b^{6} a^{15} b^{12} : a^{19} b^{18}

= \frac{1}{a ^{19} b ^{18}}

3 (- 1 + 4 i) - 2 (7 - i)

8 (5 - 3 x) - 4 (2 + 3 x) = 3

6 x = 2 x + 16

\frac{1}{6} < \frac{2 x - 19}{12} \leq \frac{2}{3}

i^{64}

What is (a^2b^3)^{-2}(a^5b^4)^{-3} ?
The solution to (a^2b^3)^{-2}(a^5b^4)^{-3} is 1/(a^{19)b^{18}}