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If x is an integer, is 9^x + 9^ -x = b?
(1) 3^x + 3^ -x = square root of (b + 2)
(2) x > 0
(1) 3^x + 3^ -x = square root of (b + 2)
(2) x > 0
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22 Apr 2007, 16:54
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Hi,
There's one problem with me in this question. I think the question should be
If x is an integer, is 9^x + 9^-x = b
1) 3^x + 3^-x = sqrt (b+2) (instead of sqrtb + 2)
2) x > 0
then, the solution can be:
9^x + 9^-x = (3^x)^2 + (3^-x)^2 = (3^x)^2 + 2.3^x.3^-x + (3^-x)^2 - 2.3^x.3^-x = (3^x + 3^-z)^2 - 2 = b
(3^x + 3^-x)^2 = b + 2
3^x + 3^-x = sqrt (b+2)
done.
There's one problem with me in this question. I think the question should be
If x is an integer, is 9^x + 9^-x = b
1) 3^x + 3^-x = sqrt (b+2) (instead of sqrtb + 2)
2) x > 0
then, the solution can be:
9^x + 9^-x = (3^x)^2 + (3^-x)^2 = (3^x)^2 + 2.3^x.3^-x + (3^-x)^2 - 2.3^x.3^-x = (3^x + 3^-z)^2 - 2 = b
(3^x + 3^-x)^2 = b + 2
3^x + 3^-x = sqrt (b+2)
done.
