can not find the proof that logarithms are the inverse of exponentials
I have been taught by the most powerful magic of mathematics - Hand waving - that the logarithms are inverses of the exponential. I have seen the graphs where each one is graph, showing they are reflective over the line $y=x$. I do not doubt they are inverses of each other; however, when I learned about the Fundamental Theorem of Calculates, I saw a proof. When I learned about the Mean Value Theorem, I saw a proof. When is learned about Riemann sums, I saw a proof. When I learned about derivatives, I saw a proof.
Where is the proof that logarithms are the inverse of exponential. I am just given a relationship $\log_b (a) = b^a$. Where's the proof?
I know there are many wise PHD here that can unleash the power of $\epsilon$ and $\delta$, crush conjectures with the squeeze theorem, follow the pigeon hole principle, make portals between fields, and even transform me into a coffee cup.
What is the proof of $\log_b(a) = b^a$?
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$\begingroup$Well, since exponential functions are 1-1 (pass the horizontal line test), and they are continuous, they must have an inverse function. The inverse is called the "logarithm". That's it! No proof needed. It is just definition. Their existence is guaranteed.
If you are trying to get your head around the "funny" way the logarithm is written, this might help.
Instead of writing $y = a^x$, write $y = exp_{a}(x)$; then $x = log_{a}(y)$. That's it. Just interchange $x$ and $y$ as you have been taught to do with other functions.
$\endgroup$ 3 $\begingroup$This proof might interest you:
Suppose we define $e^x$ to be the function $f(x)$ satisfying $f'(x) = f(x)$ with $f(0) = 1$. With implicit differentiation, we note that $$ f(f^{-1}(x)) = x \implies f'(f^{-1}(x))(f^{-1})'(x) = 1 $$ Rearranging yields $$ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} = \frac{1}{f(f^{-1}(x))} = \frac 1x $$ Now, define $f^{-1}$ to be this inverse of $f$. We deduce from the above that $f^{-1}$ has derivative $\frac 1x$. We note that since $f(0) = 1$, $f^{-1}(1) = 0$. $f^{-1}$, that is, $\ln x$, is the unique function satisfying the conditions $(f^{-1})'(x) = 1/x$ and $f^{-1}(0) = 1$. Thus, $f^{-1}$ must have all the properties of $\ln x$
We could go through the same process in reverse. The point is, once you know that $e^x$ has one of its defining properties, you may conclude that its inverse has all of the defining properties of $\ln x$. Similarly, once you know that $\ln x$ has one of its defining properties, you may conclude that its inverse has all of the defining properties of $e^x$.
I hope you find this satisfactory.
$\endgroup$ 4 $\begingroup$Let $f(x) = y = c^x$ be a exponential function, such $c>0$ and $c\neq 1$.
Take logarithms from both sides with base $c$. Then we have:
$$log_cy = log_cc^x = x$$
In other word:
$$f^{-1}(y) = g(y) = x = log_cy$$
Which means that the inverse of an exponential function is logatimic function. Also you can do the other way around by you just need to use the LHS and RHS as exponents and use $c$ as base.
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