Switching between exponential and logarithmic form

wScott

Hello you bunch of owls, I'm doing my homework at the moment and I'm curious, how woul I express the logarithmic equation

f(x) = log5 (x) + 3 in it's exponential form (where 5 is the base).

This isn't part of the homework, I'm just supposed to graph it, but I'm curious as to what the exponential form looks like.

Gib Z

[tex]5^{f(x)} = 5^{\log_5 x +3} = 5^{\log_5 x} \cdot 5^3
=125x[/tex]

wScott

Umm, I'm sure you're right, but could you elaborate on why that's correct? Could you explain how you got that i mean.

And is there a x= form of that, that's what I'vebeen trying to come up with :p.

wScott

Can anyone explain this please? Or atleast tell me what to google to find out why this works?

Gib Z

Well say I have something equal to each other. a=b.

Then x^a is equal to x^b, since a=b. So in this case, f(x)=log_5 x + 3, I did
5^(log_5 x+3) = 5^(f(x)), and I reversed the rule [tex]a^ma^n=a^{m+n}[/tex] on the 5^(log_5 x +3 ) and there we go :)

And you trying to make it equal x?
[tex]5^3x=5^{f(x)}[/tex]
[tex]x=5^{f(x)-3}[/tex]

wScott

Ahh, alright, thank you very much Sir :)

Gib Z

Thats alright, but please don't call me sir. im 15 years old lol, Sir makes me feel like im 40 >.<"

EDIT: Not that theres anything wrong with being 40 !!!:P

EDIT 2: ..OR OLDER...damn political correctness..

wScott

Could you tell me how you came up with 125x?

P.S.:haha

Gib Z

[tex]5^{f(x)} = 5^{\log_5 x +3} = 5^{\log_5 x} \cdot 5^3
[/tex]. You should be able to follow that so far. Now, by definiton of the logaritim, [tex]a^{\log_a x} =x[/tex]. And 5^3 is just 125 by expanding it..

wScott

Okay, thanks a bunch:)

d_leet

5³=125

/*extra characters*/

EDIT: I guess Gib Z beat me to it.

Gib Z

Lol just 16 minutes late d_leet :P

d_leet

Eh, it's been a long day.

HallsofIvy

log_a(x) and a^x are inverse functions.
If y= log_a(x) then x= a^y and vice-versa.

daemonakadevil

here you go with graph >

http://img123.imageshack.us/img123/6964/untitledfy8.jpg

ps: the graph would be something like that but not exactly .