How do you simplify #4^3·4^5#?

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Answer 1 · 2016-05-29T07:47:05

#4^3*4^5 = 4^8 = 65536#

For positive integer exponents we have:

#x^n = overbrace(x * x * .. * x)^"n times"#

Hence:

#x^m * x^n = overbrace(x * x * .. * x)^"m times" * overbrace(x * x * .. * x)^"n times"#

#=overbrace(x * x * .. * x)^"m + n times" = x^(m+n)#

So in our example:

#4^3*4^5 = 4^(3+5) = 4^8#

#color(white)()#
If we know our powers of #2#, then it is also helpful to use another property of exponents:

If #a, b, c > 0# then:

#(a^b)^c = a^(bc)#

So:

#4^8 = (2^2)^8 = 2^(2*8) = 2^16 = 65536#

#color(white)()#
Alternatively we could write:

#4^2 = 16#

#4^4 = 4^2 * 4^2 = 16*16 = 256#

#4^8 = 4^4 * 4^4 = 256*256 = 65536#