Question 1209333: How many pairs of integers (a,b) satisfy the equation ab^a = 648?
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
Find the prime factorization of 648:
648 = (2^3)(3^4)
The exponents in the prime factorization are 3 and 4. Since we want to write 648 in the form ab^a, a can be at most 3. Note a negative would make ab^a a negative fraction, and a=0 would make ab^a equal to zero. So a can only be 1, 2, or 3.
Try each one to see if it fits the pattern.
(1) a=1
1(b^1) = 648
b^1 = 648
b = 648
solution #1: (a,b) = (1,648)
(2) a=2
2(b^2) = 648
b^2 = 324
b = 18 or b = -18
solutions #2 and 3: (a,b) = (2,18) and (a,b) = (2,-18)
(3) a=3
3(b^3) = 648
b^3 = 216
b = 6
solution #4: (a,b) = (3,6)
ANSWER: 4 pairs of integers (a,b) satisfy ab^a = 648
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