MHB How Do You Prove the Logarithm Identity \frac{log_a n}{log_b n} = C?

[shamieh]

Given 3 constants a,b,c and 1 variable n

$$\frac{log_a n}{log_b n} = C$$ , Prove it!

I know that:

$$log_a n = \frac{log_b n}{log_b a}$$

So for easier readability I let

$$log_a n = \alpha$$
$$log_b n = \beta$$
$$log_b a = u$$

So here is what I got..

$$\alpha = \frac{\beta}{u}$$ <-- is my first formula

and this my second formula which I'm trying to solve for
$$\frac{\alpha}{\beta}$$

Subbing in for $$\alpha$$ I got:

$$\frac{\beta/u}{\beta} = \frac{1}{u} = \frac{1}{log_b a} = log_a b$$

So would that be correct?

 

[shamieh]

I also know that $$log_b b = 1$$ (according to logarithm rules)... So is there some rule where I can just change $$log_a b$$ to $$log_b b$$ to get 1/1 = 1?

 

[MarkFL]

Your expression of $$\frac{1}{u}$$ is correct, but then you switched $a$ and $b$ when you back-substituted. Why not just use the change of base formula you cited to state:

$$\frac{\log_a(n)}{\log_b(n)}=\frac{1}{\log_b(n)} \cdot\frac{\log_b(n)}{\log_b(a)}=\frac{1}{\log_b(a)}=\log_a(b)$$

I'm going to move this thread to our algebra subforum since it deals with log properties. :D

 

[shamieh]

So $$\frac{1}{log_b a} = log_a b $$ ? so, taking your advice assuming that it does equal that, what does $$log_a b = ?$$ Is that just some constant?

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Your expression of $$\frac{1}{u}$$ is correct, but then you switched $a$ and $b$ when you back-substituted. Why not just use the change of base formula you cited to state:

$$\frac{\log_a(n)}{\log_b(n)}=\frac{1}{\log_b(n)} \cdot\frac{\log_b(n)}{\log_b(a)}=\frac{1}{\log_b(a)}=\log_a(b)$$

I'm going to move this thread to our algebra subforum since it deals with log properties. :D

Thanks I edited it appropriately, I see what you're saying where I back substituted incorrectly. Now I'm just trying to figure out what exactly $$log_a b$$ is.. is it 1?

 

[shamieh]

so how is $$log_a b$$ = to some constant?are they taking $$log_a b / log_a b = log_a b * 1/log_a b = 1?$$ (thus: C) ?

 

[MarkFL]

If $a$ and $b$ are constants, then (for appropriate values) $\log_a(b)$ is just a constant as well.

And yes, the following is an identity:

$$\log_a(b)=\frac{1}{\log_b(a)}$$

Can you use the change of base theorem to prove it is true?

 

[shamieh]

If $a$ and $b$ are constants, then (for appropriate values) $\log_a(b)$ is just a constant as well.

And yes, the following is an identity:

$$\log_a(b)=\frac{1}{\log_b(a)}$$

Can you use the change of base theorem to prove it is true?

would this work?

$$log_a b = \frac{1}{log_b a}$$

$$\therefore log_a b = log_a b$$

$$\therefore \frac{log_a b}{log_a b} = 1$$

 

[MarkFL]

No, I meant something more like:

$$\log_a(b)=\frac{\log_b(b)}{\log_b(a)}= \frac{1}{\log_b(a)}$$

 

[shamieh]

Oh I see! Because log_b(b) is = 1 so we can replace it!

 

[MarkFL]

Oh I see! Because log_b(b) is = 1 so we can replace it!

Right you are! :D

 

[shamieh]

$$log_a b = \frac{log_b b}{log_b a}$$

then dividing I get

$$\frac{log_a b}{log_b b / log_b a}$$

which equals

$$log_a b * \frac{log_b a}{log_b b}$$

which = $$log_a b + log_b a $$

Which gets me no where? I'm so confused

 

[MarkFL]

How did you get that last line (the sum)?

 

[shamieh]

Because from the rules of property of logarithms $$log_a b * log_b a / 1$$ is just

$$log_a b * log_b a$$ and then when its multiplication you add right?

I just took a guess, I'm not sure what I'm doing wrong. I must be missing the big picture here though.

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Wait Oh I see now.

I think i need to change log_a b to be 1/log_b a and then they will cancel out to get 1. I didnt see that

 

[MarkFL]

I think the property you were thinking of is:

$$\log_a(bc)=\log_a(b)+\log_a(c)$$

And yes, you figured out that:

$$\log_a(b)\log_b(a)=1$$ :D

 

[shamieh]

Wow, I know those rules were allot of college algebra, but I just forgot allot of those rules. Thanks so much Mark for helping me with this proof! Took me forever to wrap my head around that