Natural logarithm question

**Bardagath** · Nov27-09, 04:44 AM

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

Sincerely,

Bardagath

**Borek** · Nov27-09, 05:33 AM

Basically you are asking why

$LaTeX Code: x = a^{\\log_a(x)}$

What is log definition?

**rasmhop** · Nov27-09, 12:14 PM

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

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. $LaTeX Code: \\log_c(a) = \\log_c(b)$ , then they must necessarily be the same (a=b) since no two different elements map to the same. Apply this to:
$LaTeX Code: \\log_a\\left(a^{\\log_a(x)}\\right) = \\log_a(x)$
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** · Nov27-09, 12:17 PM

I guess that argument does this. If we know $LaTeX Code: \\log_a(a^x) = x$ for all

, we want to use it to show $LaTeX Code: a^{\\log_a x} = x$ for all

.

**HallsofIvy** · Nov27-09, 02:11 PM

Originally Posted by g_edgar

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

There are two ways of approaching this. One is that

is defined as the inverse function to

. By the definition of "inverse functions", which requires that f(f^-1(x))= x and that f^-1(f(x))= x, then, $LaTeX Code: a^{log_a(x)}= x$ and

.

Or, just using the laws of logarithms (which are, after all, derived from the definitions),

and, if $LaTeX Code: y= a^{log_a(x)}$ , taking the

of both sides, $LaTeX Code: log_a(y)= log_a(a^{log_a(x)})$

LaTeX Code: = (log_a(x))(log_a(a))= log_a(x)

and, since

is "one-to-one" function, y= x.

**Borek** · Nov27-09, 02:58 PM

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

**Bardagath** · Nov27-09, 06:58 PM

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** · Nov27-09, 09:03 PM

Originally Posted by Borek

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** · Nov28-09, 04:39 AM

Now your LaTeX is broken