# Natural logarithm question

#### Bardagath

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

Mentor

$$x = a^{\log_a(x)}$$

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

Homework Helper
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.

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#### Borek

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

Thankyou very much for your replies!

#### HallsofIvy

Homework Helper
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

Mentor
Now your LaTeX is broken :rofl: