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**MajikWaffle****Member**- Registered: 2005-12-14
- Posts: 11

http://img228.imageshack.us/img228/1799/untitled8av.png

can someone explain this to me?

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

First rewrite them in exponential form:

9^x = a

12^x = b

16^x = a + b → 16^x = 9^x + 12^x

Solve for x.

You know, I'm not sure that's even possible without the aid of technology. My calculator says that:

x = ln[ (√(5) - 1)/2) ] / ln(3/4).

That's log base 3/4 of (√(5) - 1)/2), which is 1.67272.

Guess what? 12^1.67272 / 9^1.67272 = 63851199804262/39462211701495. More sanely, it = 1.61803.

Anyone who can solve 16^x = 9^x + 12^x for x by hand deserves a trophy.

*Last edited by ryos (2005-12-14 19:22:31)*

El que pega primero pega dos veces.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Yea, that is quite an equation, I don't believe I've ever had to solve one like it before.

My calculator comes up with 9^x = 16^(ln(3)x / 2*ln(2)) and 12^x = 16^(ln(12)x/4*ln(2))

This makes the equation:

16^x = 16^(ln(3)x / 2*ln(2)) + 16^(ln(12)x/4*ln(2))

That seems to be a step in the right direction, getting a common base. And can anyone derive 9^x = 16^... and 12^x = 16^....?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

would not b / a = (4 / 3)^x be a suitable answer?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,552

I don't know why, but in post #2 by ryos, the golden ratio

shows up twice. (√(5) - 1)/2) and 1.61803

**igloo** **myrtilles** **fourmis**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Maybe he just has the midas touch.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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