How do I find the inverse of a log function?

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Homework Help Overview

The discussion revolves around finding the inverse of logarithmic functions, specifically the function f(x) = log2 x + 3 and its variations. Participants also inquire about the domain and range of logarithmic functions.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants attempt to manipulate logarithmic expressions and question the validity of their approaches. There is a discussion on changing the base of logarithms and the implications of logarithmic identities. Some participants express uncertainty about their attempts to find inverses and the correct setup for the domain and range.

Discussion Status

Some guidance has been provided regarding the manipulation of logarithmic functions and the process of finding inverses. Participants are exploring different interpretations and methods, but there is no explicit consensus on the correct approach or solutions.

Contextual Notes

Participants note constraints related to the domain of logarithmic functions, specifically that x must be greater than zero. There is also mention of the need to avoid zero in the denominator for rational functions, which adds complexity to the problem.

hancyu
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Homework Statement



f(x) = log2 x + 3
2 log2 x − 1
how do i find the inverse of this? how do i find the range and domain of a log function?

Homework Equations

f(x) = log2 x + 3
2 log2 x − 1

is equal to
f(x) = log2 x + 3 - log2 (x − 1)2

D of f(x) = R of f-1(x)

The Attempt at a Solution



i tried changing the base but it didnt work...
 
Last edited:
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also, is the inverse of

f(x) = 2x−1 + 3

log2 (x/3) +1 = y


f(x) = log2/3(x − 2) − 4

(2/3)x+4 + 2 = y

are these correct?
 
Ok. Your attempt was sadly, incorrect, although similar to something you were probably thinking of:

\log_c ( \frac{a}{b} ) = \log_c a - \log_c b,

which is not the same as what you tried: \frac{ \log_c a}{\log_c b} = \log_c a - \log_c b, which is not true.

It might help if you let u= log_2 x so that you may view the problem easier. Doing so, solve the equation you have for you, replace the expression in x back in and solve it for x. Then swap your x and f(x), that's your inverse function!

For your second problem, not quite. Solve it for x first. So First take 3 to the other side,

2^{x-1} = f(x) - 3. After that, take log_2 of both sides, hopefully you can see the rest. Then just swap x for f(x).

The last one looks correct, good work =]
 
Presumably you know that the domain and range of any function of the form loga(x) is {x|x> 0} and all real numbers respectively.

You also should know that the domain of a rational function is all numbers such that the denominator is not zero.

Putting those together, the domain of loga(f(x))/loga(g(x)) is all x such that x is positive and g(x) is not 1 (so that log(g(x)) is not 0).
 
HallsofIvy said:
Presumably you know that the domain and range of any function of the form loga(x) is {x|x> 0} and all real numbers respectively.

You also should know that the domain of a rational function is all numbers such that the denominator is not zero.

Putting those together, the domain of loga(f(x))/loga(g(x)) is all x such that x is positive and g(x) is not 1 (so that log(g(x)) is not 0).

so the domain of the 1st one is x=>0 ? because log can never be zero or negative?

i still can't get the inverse tho...
 

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