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solving for x and y exponent

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Guest Anonymous Poster
Quote:
Original post by subflood
but I'm not sure why. I just took the hint and worked from there but I don't get why that works.

well it's 24 x 35 x 43
43 can be substituted with (22)3.
According to the arithmetic rules of powers you get 22 x 3, which equals 26.

Now following the next rule you can multiply 24 and 26 by adding the exponents to get 210.

The final equation thus is 210 x 35 = 2x x 3y.

Trivial - you don't have to do a thing, except replacing 'x' and 'y' with their literal counterparts from the left side of the equation [smile].

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Quote:
Original post by subflood
but I'm not sure why. I just took the hint and worked from there but I don't get why that works.

well it's 24 x 35 x 43
43 can be substituted with (22)3.
According to the arithmetic rules of powers you get 22 x 3, which equals 26.

Now following the next rule you can multiply 24 and 26 by adding the exponents to get 210.

The final equation thus is 210 x 35 = 2x x 3y.

Trivial - you don't have to do a thing, except replacing 'x' and 'y' with their literal counterparts from the left side of the equation [smile].

[edit] Above AP was me, too. [/edit]

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Guest Anonymous Poster
Quote:
Original post by darookie

The final equation thus is 210 x 35 = 2x x 3y.

Trivial - you don't have to do a thing, except replacing 'x' and 'y' with their literal counterparts from the left side of the equation [smile].


Actually that's wrong. The solution for x is:

x = log_2(1024 * 3^(5-y))

For example if you take y = 5 then you have the answer you arrived at: x = 10. But if you take y = 6 then x = 10 - log_2(3), for y = 7, x = log_2(1024/9) and so on.

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Original post by Anonymous Poster
Quote:
Original post by darookie

The final equation thus is 210 x 35 = 2x x 3y.

Trivial - you don't have to do a thing, except replacing 'x' and 'y' with their literal counterparts from the left side of the equation [smile].


Actually that's wrong.

No, it's not. It's a perfectly valid transformation.
Quote:

The solution for x is:
x = log_2(1024 * 3^(5-y))

For example if you take y = 5 then you have the answer you arrived at: x = 10. But if you take y = 6 then x = 10 - log_2(3), for y = 7, x = log_2(1024/9) and so on.

That was not the problem. The (or one) value of x + y was to be found. You provided a representation of x in terms of y. Typical case of thinking in the wrong direction [smile]. BTW. the interesting thing is what Charles B. pointed out - prove that 15 is a unique solution of x + y.

[edit]
Actually I think this is a textbook question that shall make readers familiar with the arithmetic rules of powers. Otherwise it would have been formulated quite differently.
[/edit]

[Edited by - darookie on October 5, 2004 7:48:33 AM]

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Guest Anonymous Poster
Quote:
Original post by darookie
... BTW. the interesting thing is what Charles B. pointed out - prove that 15 is a unique solution of x + y.

That's exactly what the previous poster was trying to say.

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Guest Anonymous Poster
(I mean that 15 is not the only solution.)

My question is: are there any complex solutions? why?

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Guest Anonymous Poster
On second thought, it's obvious that if
x = 10 + (5-y)log2(3)
then for any (real or complex) y there exists one x (also real or complex).

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Original post by Anonymous Poster
(I mean that 15 is not the only solution.)

My question is: are there any complex solutions? why?

Hint: x + y*log23 = 10 + 5*log23

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