What is the inverse of this logarithm equation?

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The discussion revolves around finding the inverse of the logarithmic equation y = -log5(-x). Initially, one participant mistakenly derived y = -5(-x) and then suggested y = log5(x) after switching x and y. However, the correct inverse is confirmed to be y = -5^(-x). The process involves manipulating the original equation and switching variables, ultimately leading to the conclusion that y = -5^(-x) is indeed the accurate inverse. Participants emphasize the importance of verifying results by substituting back into the original equation.
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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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