There is this problem that I just can not begin to solve $\displaystyle \log_{y}x=2$, $\displaystyle 5y = x + 12\log_{x}y$ this is a simultaneous equation. Can anyone help me solve this equation.

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- Nov 5th 2009, 04:56 PM #1

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- Nov 5th 2009, 05:09 PM #2

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- Nov 6th 2009, 05:03 PM #3

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- Nov 6th 2009, 06:22 PM #4

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- Nov 7th 2009, 01:44 AM #5

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- Nov 7th 2009, 01:49 AM #6
It looks like you are not very familiar with logarithms. Try to read and learn the logarithm rules (yes one must know them by heart) and solve some basic but efficient problems, and then come back and try to solve this question. It will be by far easier than it is for you now.

- Nov 7th 2009, 02:07 AM #7

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- Nov 7th 2009, 02:11 AM #8
So do I. This is why I usually refer to Wikipedia for formal lessons, and when I'm fed up with theory I can move on to the heavily-detailed examples (though sometimes I can't find any so I look for another website). And then, I can get on with the exercises in the academic department of Wikipedia when they are sufficiently reliable (let's not forget it

__is__Wikipedia ...) and when they do exist.

For example :

- Lessons : http://en.wikipedia.org/wiki/List_of...mic_identities

- Examples : http://people.hofstra.edu/Stefan_Wan...pic1/logs.html

- Exercises : [pick any suitable website for exercises]

- Nov 7th 2009, 05:28 AM #9

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so far I have tried

$\displaystyle 5(x^\frac{1}{2}) = x + 12\log_{x}(x^\frac{1}{2})$

$\displaystyle {\frac{1}{2}}\log5x= \log x +6\log_{x}x$

and then I am stock right here. Did I do it OK so far? I want to add but I don't know if they both have to be the same base to do that. Could someone please help me with the working of this so that I can use it to solve the other problem like it in the book?

- Nov 7th 2009, 05:34 AM #10

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It's far simpler than you're making it out to be:

By the definition of a logarithm, $\displaystyle log_x(x)=1$.

Why? -- we know that $\displaystyle x^{log_x(a)} = a$. So $\displaystyle x^{log_x(x)} = x$ and therefore $\displaystyle log_x(x)=1$.

Using this and the fact that $\displaystyle log_b(a^n) = n\cdot log_b(a)$ for any $\displaystyle a,b,n$, we get:

$\displaystyle 5\sqrt{x} =x + 12log_x(x^{\frac{1}{2}}) \Rightarrow 5\sqrt{x} = x + 12\cdot \frac{1}{2} \cdot \log_x(x) \Rightarrow $ $\displaystyle 5\sqrt{x} = x + 6 \Rightarrow 25x = x^2 + 12x + 36 \Rightarrow x^2-13x+36=0$

I believe you can solve that.

- Nov 7th 2009, 06:54 AM #11

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- Nov 7th 2009, 07:01 AM #12

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