How do you simplify #(7^4)^3#?

2 Answers
Jan 7, 2017

#7^(12)#

Explanation:

Using the #color(blue)"law of exponents"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(a^mxxa^n=a^(m+n))color(white)(2/2)|)))#
which can be extended to the product of more than 2 terms.

#rArr(7^4)^3=7^4xx7^4xx7^4#

#=7^(4+4+4)=7^(12)#

Note that the exponent 12, is also obtained by #color(blue)"multiplying"# 3 and 4 together. This leads us to a further law of exponents.

#"That is " color(red)(bar(ul(|color(white)(2/2)color(black)((a^m)^n=a^(mn))color(white)(2/2)|)))#

#rArr(7^4)^3=7^(4xx3)=7^(12)#

Jan 7, 2017

#(7^4)^3 = 7^12#

Explanation:

Note that if #a# is any number and #m, n# are positive integers then:

#(a^m)^n=overbrace(a^mxxa^mxx...xxa^m)^"n times"#

#color(white)((a^m)^n)=overbrace(overbrace(axxaxx...xxa)^"m times"xxoverbrace(axxaxx...xxa)^"m times"xx...xxoverbrace(axxaxx...xxa)^"m times")^"n times"#

#color(white)((a^m)^n)=overbrace(axxaxx...xxa)^"mn times"#

#color(white)((a^m)^n)=a^(mn)#

So in our example:

#(7^4)^3 = 7^(4*3) = 7^12#