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Question about Log Laws

  • Thread starter escryan
  • Start date
13
0
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.
 

Answers and Replies

Dick
Science Advisor
Homework Helper
26,258
618
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.
 
1,631
4
well, by definition of log we have

[tex]log_a(x)=b<=> a^b=x[/tex]

Now lets substitute [tex] b=log_a(x)[/tex] in
[tex]a^b=x[/tex] So:

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

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

[tex]fg(x)=f(g(x))=x=>a^{log_ax}=x[/tex] and also

[tex] g(f(x))=log_a(a^x)=x[/tex]

Edit: Dick was faster!
 
Dick
Science Advisor
Homework Helper
26,258
618
well, by definition of log we have

[tex]log_a(x)=b<=> a^b=x[/tex]

Now lets substitute [tex] b=log_a(x)[/tex] in
[tex]a^b=x[/tex] So:

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

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

[tex]fg(x)=f(g(x))=x=>a^{log_ax}=x[/tex] and also

[tex] g(f(x))=log_a(a^x)=x[/tex]

Edit: Dick was faster!
You are slow because you write more. Doesn't mean you think slower. I appreciate the TeX though.
 
Last edited:
13
0
Oh I see now! Thanks so much for your help Dick and sutupidmath!
 

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