Converting a log equation to exponential equation

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The discussion revolves around converting the logarithmic equation f(x) = log5(x) + 3 into its exponential form. The exponential equivalent is derived as x = 5^(y - 3), which can also be expressed as x = (1/125) * 5^y. Participants clarify that the logarithmic function can be graphed, but the focus is on finding the correct exponential representation. The conversation confirms that the topic fits within the general math section. The transformation from logarithmic to exponential form is essential for understanding the relationship between these functions.
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Homework Statement


the question asks to graph the equation:
(don't know how to use latex sorry)
f(x) = log5 (x) + 3 where the 5 is the base
Just for my curiosity what would the exponential equation be?
I can graph it, just can't get the exponential form
Tell me if this should go in the general math section.
 
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You already put it in the general math section lol, look there for the working.
5^{f(x)} = 125x
 
Can that be expressed as x=? and can you show me how you did that?
 
wScott said:

Homework Statement


the question asks to graph the equation:
(don't know how to use latex sorry)
f(x) = log5 (x) + 3 where the 5 is the base
Just for my curiosity what would the exponential equation be?
I can graph it, just can't get the exponential form
Tell me if this should go in the general math section.

"Solve" for x: y- 3= log5(x) so, using the definition of log5(x) as the inverse of 5x, x= 5y-3= (1/125)5y.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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