math help plz?
回答 (8)
2017-12-24 11:15 am
✔ 最佳答案
1296 = 2^4 X 3^4 = 6^4(-6)^4 also works.
And you could also make that (6^2)^2 = 36^2
Or (-36)^2
And lastly there is the trivial 1296^1
So all together, there are 5 pairs of integers that will work. Integers can be positive or negative. I'm surprised so many fine people are missing the negative possibilities for a.
2017-12-24 12:33 pm
a^b
= 1296
= 2^4 × 3^4
= 1296
= 2^4 × 3^4
2017-12-24 12:31 pm
a^b=1296
b*ln(a) = ln(1296)
b = ln(1296)/ln(a)
b = ln(6^4)/ln(a)
a must be 6, 6^2, 6^4, 1/6, 1/6^2, 1/6^4
b*ln(a) = ln(1296)
b = ln(1296)/ln(a)
b = ln(6^4)/ln(a)
a must be 6, 6^2, 6^4, 1/6, 1/6^2, 1/6^4
2017-12-24 11:11 am
3:
1296^1
6^4
36^2
1296^1
6^4
36^2
2017-12-24 12:07 pm
Factors of 1296 = 2^4*3^4
Square root is 36, so stop at 36 because anything above that will have been included
1,1296
2,648
3,432
4,324
6,216
8,162
9,144
12,108
16,81
18,72
24,56
27,48
36,36
13 pairs
Square root is 36, so stop at 36 because anything above that will have been included
1,1296
2,648
3,432
4,324
6,216
8,162
9,144
12,108
16,81
18,72
24,56
27,48
36,36
13 pairs
2017-12-24 10:51 am
1296 = 36^2 or 6^4.
That's it. 2 pairs.
That's it. 2 pairs.
2017-12-24 10:20 am
1296^1 = 1296.
Prime factorization of 1296 is
2*648 = 2*2*324 = 2*2*2*162 = 2^4*81 = 2^4*3^4.
Therefore,
6^4 = 1296 and 36^2 = 1296.
That's the end of it. Three pairs of integers.
Prime factorization of 1296 is
2*648 = 2*2*324 = 2*2*2*162 = 2^4*81 = 2^4*3^4.
Therefore,
6^4 = 1296 and 36^2 = 1296.
That's the end of it. Three pairs of integers.
2017-12-24 10:13 am
Well certainly 1296^1 = 1296. That's the highest possible value for a.
And the lowest possible value a could take is 2 (because 1^integer will always be 1). Let's see what b would be for a = 2:
2^b = 1296
b log 2 = log 1296
b = log 1296/log 2 = 10.34, not an integer. But now we know that b is at most 10. So just try all the values of b from 1 to 10 and see if there are any more solutions. For instance, if b = 10
a^10 = 1296
a = 1296^(1/10) = 2.047, not an integer.
Repeat with 1296^(1/9), 1296^(1/8), etc.
And the lowest possible value a could take is 2 (because 1^integer will always be 1). Let's see what b would be for a = 2:
2^b = 1296
b log 2 = log 1296
b = log 1296/log 2 = 10.34, not an integer. But now we know that b is at most 10. So just try all the values of b from 1 to 10 and see if there are any more solutions. For instance, if b = 10
a^10 = 1296
a = 1296^(1/10) = 2.047, not an integer.
Repeat with 1296^(1/9), 1296^(1/8), etc.
收錄日期: 2021-04-18 18:03:53
原文連結 [永久失效]:
https://answers.yahoo.com/question/index?qid=20171224020921AAd5mU2