Flash Card Question - NP #14

[jessica.berrada]

The question asks: (6^4)(50^3)/ (2^4)(3^4)(10^3)

My strategy was to break the larger number into primes and got the following:
((2^4)(3^4))x(2^3)x(5^2)^3 / (2^4)(3^4) x (2^3)(5^3)

I eliminated the 2^4, 3^4, and 2^3 in the numerator and denominator. I then eliminated the 5^3 in the numerator and denominator and was left with 5^2 in the numerator for a final answer of 25.

However, the flashcard answer was 5^3 or 125. The flashcard strategy cancelled the 6^4 in the numerator with the 2^4 and 3^4 in the denominator and was left with 50^3/10^3 = 5^3. But I thought when you divide exponential terms you had to subtract the exponents. So, would it be 50/10 = 5^3-3? Also, why wouldn't you break the 50^3 and 10^3 down into their prime factors?

[RonPurewal]

hi,

I then eliminated the 5^3 in the numerator and denominator and was left with 5^2 in the numerator for a final answer of 25.

the other eliminations seem to be fine.
the numerator contains (5^2)^3 = 5^6; the denominator contains 5^3.
when you divide 5^6 by 5^3, the result is 5^3, not 5^2. (if you know that these exponents are subtracted, there you go. if not, you can always just write out 5^6/5^3 = (5x5x5x5x5x5)/(5x5x5) and then cross out three pairs of 5's)

I thought when you divide exponential terms you had to subtract the exponents. So, would it be 50/10 = 5^3-3?

that rule only applies when the base number (the number that's raised to the power) is the same in both.
so, for instance, 7^5 divided by 7^3 is 7^(5 - 3) = 7^2, because the base number 7 is the same in both.

Also, why wouldn't you break the 50^3 and 10^3 down into their prime factors?

you can do that (and probably should, if you are unfamiliar with the other exponent rules). you just don't have to, since 10 goes evenly into 50.