Logarithmic Simplification: Understanding the Natural Logarithm

[Bardagath]

Hello,

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

log_a(a^log_a(x)) = log_a(x) ==> a^log_a(x) = x

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



Bardagath

 

[Borek]

Basically you are asking why

x = a^{\log_a(x)}

What is log definition?

 

[rasmhop]

Do you know the rules:
\log_b(a^x) = x\log_b(a)
\log_b(b) = 1
If you do, then it follows from first applying the first rules and then seeing:
\log_a(x) \log_a(a) = \log_a(x) \times 1 = \log_a(x)

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. \log_c(a) = \log_c(b), then they must necessarily be the same (a=b) since no two different elements map to the same. Apply this to:
\log_a\left(a^{\log_a(x)}\right) = \log_a(x)
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).

 

[g_edgar]

I guess that argument does this. If we know \log_a(a^x) = x for all x, we want to use it to show a^{\log_a x} = x for all x.

 

[HallsofIvy]

I guess that argument does this. If we know \log_a(a^x) = x for all x, we want to use it to show a^{\log_a x} = x for all x.

There are two ways of approaching this. One is that log_a(x) is defined as the inverse function to a^x. By the definition of "inverse functions", which requires that f(f^-1(x))= x and that f^-1(f(x))= x, then, a^{log_a(x)}= x and log_a(a^x)= x.

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

 

[Borek]

Obvious typo - you meant a^, not e^ :)

 

[Bardagath]

I don't know why it didn't fall into place earlier but I woke up today and it fits;

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

Thankyou very much for your replies!

 

[HallsofIvy]

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]

Now your LaTeX is broken