How do you simplify 4^3·4^5?
1 Answer
May 29, 2016
Explanation:
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
If we know our powers of
If
(a^b)^c = a^(bc)
So:
4^8 = (2^2)^8 = 2^(2*8) = 2^16 = 65536
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