How do you simplify #(b^3 a^5)/(b^2 a^4)#?

2 Answers
Apr 10, 2016

Answer:

#(b^3a^5)/(b^2a^4)=ba#

Explanation:

#(b^3a^5)/(b^2a^4)=(b*b*b*a*a*a*a*a)/(b*b*a*a*a*a)#

So you should cancel b's and a's in nominator and denominator. So the result would be #b*a#.

More scientifically:

#(b^3a^5)/(b^2a^4)= b^(3-2) * a^(5-4) = b^1 * a^1 = b*a #

Apr 10, 2016

Answer:

ab

Explanation:

Using the following #color(blue)" rule of exponents " #

#color(red)(|bar(ul(color(white)(a/a)color(black)( a^m/a^n hArr a^(m-n))color(white)(a/a)|)))#

#rArr (b^3a^5)/(b^2a^4) = b^3/b^2xxa^5/a^4 = b^(3-2)xxa^(5-4) = ab#