Which exponential property do you use first to simplify $\frac{(2a^2bc^2)(6abc^3)}{4ab^2c}$ ?

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Answer 1 · 2014-11-05T10:38:34

I would use the exponential property

$b^x cdot b^y=b^{x+y}$

to simplify the numerator first.

${(2a^2bc^2)(6abc^3)}/{4ab^2c}={12a^3b^2c^5}/{4ab^2c}$

I hope that this was helpful.

Which exponential property do you use first to simplify \frac{(2a^2bc^2)(6abc^3)}{4ab^2c}\frac{(2a^2bc^2)(6abc^3)}{4ab^2c}?