SOLUTION: how do you solve 12^9 over 12^2 how do you solve (6^2)^4 in simplest form

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Question 776496: how do you solve 12^9 over 12^2
how do you solve (6^2)^4 in simplest form

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+12%5E9+%2F+12%5E2+
Follow the general rule:
+a%5Ec+%2F+a%5Eb+=+a%5E%28+c-b%29++
+12%5E9+%2F+12%5E2+=+12%5E%28+9-2+%29+
+12%5E%28+9-2+%29+=+12%5E7+
--------------------
+%28+6%5E2%29%5E4+
Think of this as:
+6%5E2%2A6%5E2%2A6%5E2%2A6%5E2+
According to the above rule, this equals
+6%5E%28+2+%2B+2+%2B+2+%2B+2+%29+
+6%5E8+
So, you just multiply the exponents
If you had +%286%5E10%29%5E10+ that would equal
+6%5E100+
Or, as I showed,
+6%5E%28+10+%2B+10+%2B+10+%2B+10+%2B+10+%2B+10+%2B+10+%2B+10+%2B+10+%2B+10+%29+