What is the inverse of this logarithm equation?

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    Inverse Logarithm
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Homework Help Overview

The discussion revolves around finding the inverse of the logarithmic equation y = -log5(-x). Participants explore various methods to derive the inverse and question the correctness of their results.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the inverse by switching x and y and expresses uncertainty about their results. Some participants provide feedback on the correctness of the results and suggest checking the original equation for validation.

Discussion Status

Participants are actively discussing the correctness of different results for the inverse. There is a mix of confirmations and corrections regarding the derived expressions, indicating an ongoing exploration of the topic.

Contextual Notes

Some participants reference the method of switching coordinates from a table of values, while others emphasize checking the original equation to verify results. There is a noted confusion about the signs and forms of the expressions derived.

supernova1203
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Homework Statement



What is the inverse of this logarithm equation?

y=-log5(-x)

i tried it and i got y=-5(-x)hm.. you know how people say that if you want to find the inverse graph of something just switch the x and y coordinates from the table of values? Well i also tried that approach and apparently the inverse is y=log5x

atleast that's how it looks on table of values and on the graph...i don't know why i went with the y=-5(-x)I don't know if i got it right or not, someone kind enough to take a look and see if i got it right or not? or maybe just give me the solution outright so i know if i did it correctly or not? :P
 
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Your result is correct. If you start from your original equation, multiply both sides by negative 1, you can easily switch x and y, and carry couple simple steps to find y as a function of x.
 
hm... so y=log5x is correct?
 
supernova1203 said:
hm... so y=log5x is correct?

No! Your first result was correct. y=-5^(-x)
 
ahh ok ty
 
Original equation: y=-log_{5}(-x)

-y=log_{5}(-x)
Now express exponential form :
5^{-y}=-x

Now, switch x and y to create inverse:
5^{-x}=-y,

and then simply, y=-5^{-x}
 
supernova1203 said:

Homework Statement



What is the inverse of this logarithm equation?

y=-log5(-x)




i tried it and i got y=-5(-x)


hm.. you know how people say that if you want to find the inverse graph of something just switch the x and y coordinates from the table of values? Well i also tried that approach and apparently the inverse is y=log5x

atleast that's how it looks on table of values and on the graph...i don't know why i went with the y=-5(-x)


I don't know if i got it right or not, someone kind enough to take a look and see if i got it right or not? or maybe just give me the solution outright so i know if i did it correctly or not? :P

To check if you're right, when you get to the equation x=-5^{-y}, just plug that value of x into your original equation y=-\log_5(-x) and see if you get an equality.
 

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