What is (5x^{3n+1}y^{4n+2})(3x^{6n-2}y^{3n+4}) ?

The solution to (5x^{3n+1}y^{4n+2})(3x^{6n-2}y^{3n+4}) is 15x^{9n-1}y^{7n+6}

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simplify (5x^{3n+1}y^{4n+2})(3x^{6n-2}y^{3n+4})

simplify

(5 x^{3 n + 1} y^{4 n + 2}) (3 x^{6 n - 2} y^{3 n + 4})

15 x^{9 n - 1} y^{7 n + 6}

Solution steps

(5 x^{3 n + 1} y^{4 n + 2}) (3 x^{6 n - 2} y^{3 n + 4})

Apply exponent rule:

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

x^{3 n + 1} x^{6 n - 2} = x^{3 n + 1 + 6 n - 2}

= 5 y^{4 n + 2} \cdot 3 x^{3 n + 1 + 6 n - 2} y^{3 n + 4}

Apply exponent rule:

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

y^{4 n + 2} y^{3 n + 4} = y^{4 n + 2 + 3 n + 4}

= 5 \cdot 3 x^{3 n + 1 + 6 n - 2} y^{4 n + 2 + 3 n + 4}

Simplify

3 n + 1 + 6 n - 2 : 9 n - 1

= 5 \cdot 3 x^{9 n - 1} y^{4 n + 3 n + 2 + 4}

Simplify

4 n + 2 + 3 n + 4 : 7 n + 6

= 5 \cdot 3 x^{9 n - 1} y^{7 n + 6}

Multiply the numbers:

5 \cdot 3 = 15

= 15 x^{9 n - 1} y^{7 n + 6}

9.35 \cdot 1 0^{- 3}

5 e^{- 4 ∣ t ∣}

e^{- \frac{21}{10}}

2 15 + 4

What is (5x^{3n+1}y^{4n+2})(3x^{6n-2}y^{3n+4}) ?
The solution to (5x^{3n+1}y^{4n+2})(3x^{6n-2}y^{3n+4}) is 15x^{9n-1}y^{7n+6}