Switching between exponential and logarithmic form

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SUMMARY

The discussion centers on converting the logarithmic equation f(x) = log5(x) + 3 into its exponential form. The correct transformation is expressed as 5f(x) = 125x, derived from the properties of logarithms and exponents. Participants clarify that 5f(x) equals 5log5(x) + 3, which simplifies to 5log5(x) * 53 = 125x. The inverse relationship between logarithmic and exponential functions is emphasized, confirming that if y = loga(x), then x = ay.

PREREQUISITES
  • Understanding of logarithmic functions, specifically log5(x).
  • Familiarity with exponential functions and their properties.
  • Knowledge of inverse functions, particularly the relationship between logarithms and exponentials.
  • Basic algebra skills for manipulating equations.
NEXT STEPS
  • Study the properties of logarithms and exponents in detail.
  • Learn how to graph logarithmic and exponential functions effectively.
  • Explore the concept of inverse functions in mathematics.
  • Practice converting various logarithmic equations into their exponential forms.
USEFUL FOR

Students learning algebra, particularly those studying logarithmic and exponential functions, as well as educators seeking to explain these concepts effectively.

wScott
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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.
 
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5^{f(x)} = 5^{\log_5 x +3} = 5^{\log_5 x} \cdot 5^3<br /> =125x
 
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.
 
Can anyone explain this please? Or atleast tell me what to google to find out why this works?
 
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 a^ma^n=a^{m+n} on the 5^(log_5 x +3 ) and there we go :)

And you trying to make it equal x?
5^3x=5^{f(x)}
x=5^{f(x)-3}
 
Ahh, alright, thank you very much Sir :)
 
Thats alright, but please don't call me sir. I am 15 years old lol, Sir makes me feel like I am 40 >.<"

EDIT: Not that there's anything wrong with being 40 !:P

EDIT 2: ..OR OLDER...damn political correctness..
 
Last edited:
Could you tell me how you came up with 125x?

P.S.:haha
 
Last edited:
5^{f(x)} = 5^{\log_5 x +3} = 5^{\log_5 x} \cdot 5^3<br />. You should be able to follow that so far. Now, by definition of the logaritim, a^{\log_a x} =x. And 5^3 is just 125 by expanding it..
 
  • #10
Okay, thanks a bunch:)
 
  • #11
wScott said:
Could you tell me how you came up with 125x?

P.S.:haha

53=125

/*extra characters*/

EDIT: I guess Gib Z beat me to it.
 
  • #12
Lol just 16 minutes late d_leet :P
 
  • #13
Gib Z said:
Lol just 16 minutes late d_leet :P

Eh, it's been a long day.
 
  • #14
loga(x) and ax are inverse functions.
If y= loga(x) then x= ay and vice-versa.
 
  • #15
here you go with graph >

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

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

:zzz:
 
Last edited by a moderator:

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