/sci/ - Science & Math

Nope. There is no tuple (a,b,c) such that this identity holds for all x>0. It is relatively easy to check by doing the following proof by contradiction:

<span class="math">\log_c b = \log_a(\log_b b) = \log_a 1 = 0[/spoiler]

Therefore <span class="math">\frac{\log b}{\log c}=0[/spoiler], thus <span class="math">b=1[/spoiler], which is a contradiction because <span class="math">\log_1(x)[/spoiler] is undefined for <span class="math">x\neq 1[/spoiler].

>>5699143

Good answer, if I'm following it correctly, thanks.

ln(ln(e)) = log base x of (e) --> 0 = log base x of (e) --> undefined

not a real proof, but it doesn't work in this case, which implies there's no general rule

>>5699187
You're welcome.

In general, when attempting to show that something can be written in terms of a given function (here in terms of <span class="math">\log_c x[/spoiler]), the strategy is:

- Check if it looks like it's possible on a trivial example (e.g. take a=b=1 and try to find c),
- In the above case, it's impossible. You can either figure it out and then try to prove it, or start proving it right away, because the same kind of idea is used:
- The idea is, <span class="math">\log_c(x)[/spoiler] only depends on a single parameter <span class="math">c[/spoiler]. In general, there could be a few parameters, but usually when you try to write something "in terms of" something, you don't have that many parameters (but you could have something like <span class="math">\lambda e^{\alpha x}[/spoiler] with two parameters).
- Because it depends on only one parameter, you don't have a lot of leeway for that parameter <span class="math">c[/spoiler] if you start fixing values of <span class="math">x[/spoiler]. If you choose your values that are "independent" in some sense, each value of <span class="math">x[/spoiler] that you observe gives you an equation on <span class="math">c[/spoiler]. Here I observed the value <span class="math">x=b[/spoiler], and it gave me <span class="math">\log_c b=0[/spoiler].
- Once you have more independent equations than you have parameters, you've usually won: all of them cannot be satisfied at once.

If I ask you why you can't write <span class="math">x^2[/spoiler] as <span class="math">ax+b[/spoiler],
1) x=0 gives <span class="math">0^2=a\times 0+b[/spoiler] so b=0,
2) x=1 gives <span class="math">1^2=a\times 1+b=a[/spoiler] so a=1 (using b=0),
2) x=-1 gives <span class="math">(-1)^2=a\times (-1)+b=-a[/spoiler] so a=-1, which is a contradiction.

Get the idea?

, <span class="math">x=1[/spoiler] and <span class="math">x=-1[/spoiler].

>>5699113
Yes. When a = b = c = 1.

>>5699331
> Divided by zero

http://www.wolframalpha.com/input/?i=log%28a%2C+log%28b%2C+x%29%29%3Dlog%28y%2C+x%29

http://www.wolframalpha.com/input/?i=solve+log%28a%2C+log%28b%2C+x%29%29%3Dlog%28c%2C+x%29+for+c

>>5699612

The fact that x shows up in the solution for c indicates that c doesn't depend on only a and b.

>>5699883
You would have to take 2 different values of x and prove that it yields 2 different results, though (or to plot it as a function of x and see that it's not a constant).

Unlucky things happen: sometimes you want to see that something depends on x, and you get an expression that involves x, but it actually simplifies to a constant. It is relatively rare when using tools like WA, but they are not all-powerful.

>>5700428

Well, Wolfram usually simplifies things as much as possible, but yeah, I should have added that disclaimer