48 = 2∙24 = 2∙2∙12 = 2∙2∙2∙6 = 2∙2∙2∙2∙3
12 = 2∙6 = 2∙2∙3
y13 = y∙y∙y∙y∙y∙y∙y∙y∙y∙y∙y∙y∙y
y7 = y∙y∙y∙y∙y∙y∙y
Cancel the 2 factors of 2 in the bottom into 2 of the factors
of 2 in the numerator:
Cancel the 7 factors of y in the bottom into 7 of the factors
of y in the numerator:
And all that's left is
and since 2∙2 = 4, and y∙y∙y∙y∙y∙y = y6 <-- final answer.
There is a much shorter way that involves subtracting the
exponents to see how many factors would cancel if you did
it this long way. See if you can figure out the short way
by yourself, so you can do problems like this is just one
or two steps.
Edwin
