Thread:

1. log Problem

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

2. Originally Posted by scrible
There is this problem that I just can not begin to solve $\log_{y}x=2$, $5y = x + 12\log_{x}y$ this is a simultaneous equation. Can anyone help me solve this equation.
Notice that $\log_{y}(x)=2 \Rightarrow y^2=x$

Substitute $y=x^{\frac{1}{2}}$ into the second equation.

3. Well to continue this problem I am having trouble working out $5(x^\frac{1}{2})=x + 12\log_{x}(x^\frac{1}{2})$ can some one please help me with this.

4. Originally Posted by scrible
Well to continue this problem I am having trouble working out $5(x^\frac{1}{2})=x + 12\log_{x}(x^\frac{1}{2})$ can some one please help me with this.
Can you see why $log_x(x^{\frac{1}{2}}) = \frac{1}{2}$?

After you do that, you should be able to solve the rest.

5. Originally Posted by Defunkt
Can you see why $log_x(x^{\frac{1}{2}}) = \frac{1}{2}$?

After you do that, you should be able to solve the rest.

In the problem do I have to change the base?

6. Originally Posted by scrible
In the problem do I have to change the base?
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.

7. Originally Posted by Bacterius
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.
Do you know of any sights I can go to with examples like these? The book is not explaining it well enough for me. I learn by examples.

8. Originally Posted by scrible
Do you know of any sights I can go to with examples like these? The book is not explaining it well enough for me. I learn by examples.
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]

9. Originally Posted by Bacterius
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 : List of logarithmic identities - Wikipedia, the free encyclopedia
- Examples : Properties of Logarithms
- Exercises : [pick any suitable website for exercises]
so far I have tried

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

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

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?

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

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

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

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

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

I believe you can solve that.

11. Originally Posted by Defunkt
It's far simpler than you're making it out to be:

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

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

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

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

I belive you can solve that.
Thanks a million. You know I keep going over the log laws, it is the application of the laws that is killing me. Do you know of any web sight with tricky examples like these?

12. Originally Posted by scrible
Thanks a million. You know I keep going over the log laws, it is the application of the laws that is killing me. Do you know of any web sight with tricky examples like these?
I don't, sorry, but you should try looking for some on google.
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