A Third function

Ralphz

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For a=b(exp)c, fixing the b, or the base, yields the exponential function or its inverse, the logarithmic function; fixing the exponent, or c, yields the power function. So what function do you get when you fix a?
 

topsquark

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[math]a = b^c[/math]
Let's use base e and just call it "log."

[math]log(a) = c ~ log(b)[/math]
[math]c = \dfrac{log(a)}{log(b)} = \dfrac{ \text{constant} }{log(b)}[/math]
so c as a function of b is essentially just 1 over log. Nothing new.

-Dan

Addendum: Notation comment. What you wrote "a = b(exp)c" is written in text as "a = b^c"
 

Ralphz

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Thank you. I am think what happens when the base, b, varies continuously and the exponent varies continuously while the argument is held constant.
 

Ralphz

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To put it another way, what is the function in which the base is the dependent variable and the exponent then varies continuously as the argument is held constant. The inverse function would be the exponent as dependent variable as the base varies continuously with the argument constant.
 

JeffM

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To put it another way, what is the function in which the base is the dependent variable and the exponent then varies continuously as the argument is held constant. The inverse function would be the exponent as dependent variable as the base varies continuously with the argument constant.
This is called an implicit function. If a is a positive constant

[MATH]a =x^y \implies x = a^{1/y} \text { AND }y = a * \dfrac{ln(a)}{ln(x)} \text { unless } a = 1.[/MATH]
It is your choice.
 

Ralphz

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I thank you all for your clear presentations. As I look at these functions, I would like to ask the following:
The above is one of 3 potential pairs of variables; the other 2 represent the power and exponential functions, both deeply central to mathematics ordained as such by their respective names. What do we understand of the role of the 3rd pairing? What are its correspondences in nature and to deeper math? Or perhaps it is just the odd man out?
 
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