Natural logarithm question

  • Thread starter Bardagath
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10
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Hello,

I made a mistake in the title of this thread and this question is on general logarithms;

loga(aloga(x)) = loga(x) ==> aloga(x) = x

Can someone enlighten me on why loga(aloga(x)) simplifies to loga(x)? Can someone prove why this is true? Futhermore, why does this imply that aloga(x) = x? I am having trouble getting my head around this

Sincerely,

Bardagath
 

Borek

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Basically you are asking why

[tex]x = a^{\log_a(x)}[/tex]

What is log definition?
 
424
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Do you know the rules:
[tex]\log_b(a^x) = x\log_b(a)[/tex]
[tex]\log_b(b) = 1[/tex]
If you do, then it follows from first applying the first rules and then seeing:
[tex]\log_a(x) \log_a(a) = \log_a(x) \times 1 = \log_a(x)[/tex]

The logarithm is what we call an injective function (also called one-to-one function I believe) which basically means that if two elements a and b are mapped to the same element, i.e. [itex]\log_c(a) = \log_c(b)[/itex], then they must necessarily be the same (a=b) since no two different elements map to the same. Apply this to:
[tex]\log_a\left(a^{\log_a(x)}\right) = \log_a(x)[/tex]
if you have already shown that log is injective (otherwise you need to use some other property, but the argument seems to suggest that this is the property used).
 
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I guess that argument does this. If we know [itex]\log_a(a^x) = x[/itex] for all [itex]x[/itex], we want to use it to show [itex]a^{\log_a x} = x[/itex] for all [itex]x[/itex].
 

HallsofIvy

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I guess that argument does this. If we know [itex]\log_a(a^x) = x[/itex] for all [itex]x[/itex], we want to use it to show [itex]a^{\log_a x} = x[/itex] for all [itex]x[/itex].
There are two ways of approaching this. One is that [itex]log_a(x)[/itex] is defined as the inverse function to [itex]a^x[/itex]. By the definition of "inverse functions", which requires that f(f-1(x))= x and that f-1(f(x))= x, then, [itex]a^{log_a(x)}= x[/itex] and [itex]log_a(a^x)= x[/itex].

Or, just using the laws of logarithms (which are, after all, derived from the definitions), [itex]log_a(a^x)= x log_a(a)= x[/itex] and, if [itex]y= a^{log_a(x)}[/itex], taking the [itex]log_a[/itex] of both sides, [itex]log_a(y)= log_a(a^{log_a(x)})[/itex][itex]= (log_a(x))(log_a(a))= log_a(x)[/itex] and, since [itex]log_a[/itex] is "one-to-one" function, y= x.
 
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Borek

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Obvious typo - you meant a^, not e^ :)
 
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I don't know why it didn't fall into place earlier but I woke up today and it fits;

Yes, loga(x)loga(a) = loga(x) . 1 = loga(x)

Thankyou very much for your replies!
 

HallsofIvy

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Obvious typo - you meant a^, not e^ :)
Yes, of course. Thanks. I have edited my post so I can pretend I didn't make that mistake.
 

Borek

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Now your LaTeX is broken :rofl:
 

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