Uncovering the Mysteries of Log Laws: Exploring the Equations e^lnx and 8^log8x

[escryan]

I don't know how I managed to forget this one, but I did somehow...

If there's something like:

e^lnx, why is that equal to just x?

and same goes for sokmething like:

8^log8x which is just equal to x.

I'm just wondering how, algebraically, one could show this to be true.

 

[Dick]

You generally DEFINE ln(x) to be the inverse function of e^x. Or vice versa depending on which you define first. So you don't show it algebraically, it largely a matter of definition.

 

[sutupidmath]

well, by definition of log we have

$$log_a(x)=b<=> a^b=x$$

Now let's substitute $$b=log_a(x)$$ in
$$a^b=x$$ So:

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

Or, since $$f(x)=a^x$$ and $$g(x)=log_ax$$ are inverse functions, so it means that they cancel each other out. That is

$$fg(x)=f(g(x))=x=>a^{log_ax}=x$$ and also

$$g(f(x))=log_a(a^x)=x$$

Edit: Dick was faster!

 

[Dick]

well, by definition of log we have

$$log_a(x)=b<=> a^b=x$$

Now let's substitute $$b=log_a(x)$$ in
$$a^b=x$$ So:

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

Or, since $$f(x)=a^x$$ and $$g(x)=log_ax$$ are inverse functions, so it means that they cancel each other out. That is

$$fg(x)=f(g(x))=x=>a^{log_ax}=x$$ and also

$$g(f(x))=log_a(a^x)=x$$

Edit: Dick was faster!

You are slow because you write more. Doesn't mean you think slower. I appreciate the TeX though.

 

[escryan]

Oh I see now! Thanks so much for your help Dick and sutupidmath!