How do you simplify #b^ { 4} \cdot b ^ { 7} \cdot b ^ { 5}#?

2 Answers
Jan 15, 2018

See a solution process below:

Explanation:

Use this rule for exponents to simplify the expression:

#x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))#

#b^color(red)(4) xx b^color(blue)(7) xx b^color(green)(5) => b^(color(red)(4) + color(blue)(7) + color(green)(5)) => b^16#

Jan 15, 2018

This is a law of exponents
#a^mxxa^n=a^(m+n)#
Similarly
#a^mxxa^nxxa^p=a^(m+n+p)#
#a# here in the problem #b#
and #m,n,p# are #4,7,5#
So....
#b^4xxb^7xxb^5=b^(4+7+5)#
Which gives
#b^16#