December 11th, 2017, 08:18 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,444 Thanks: 106  lnx and e^x
$\displaystyle \ln x$ and e$\displaystyle ^{x}$ are inverse functions: $\displaystyle e^{\ln x}=x$ and $\displaystyle \ln e^{x}=x$ But what are $\displaystyle \ln x$ and $\displaystyle e^{x}$? Options: 1) Define $\displaystyle e^{x}$ and $\displaystyle \ln x$ independently: $\displaystyle e^{x} = 1+{}\sum_{n=1}^{\infty}\frac{x^{n}}{n!}$ $\displaystyle \ln x=\int_{1}^{x}\frac{dx}{x}$ Then show they are inverses and derive the usual properties. 2) Choose either series or integral definition as fundamental, and define the other as inverse. If you choose integral definition, the series for $\displaystyle e^{x}$ is easily derived: By definition $\displaystyle e^{x}$ is the inverse function of $\displaystyle \ln x$. If $\displaystyle y=\ln^{1}x. x=\ln y$ and $\displaystyle dx=\frac{1}{y}dy$ so that $\displaystyle \frac{dy}{dx}= y =\ln^{1}(x)$, and $\displaystyle \frac{d^{2}y}{dx^{2}}=\ln^{1}(x)$,..... In general $\displaystyle \frac{d^{n}y}{dx^{n}}=\ln^{1}(x)$ and $\displaystyle \frac{d^{n}y}{dx^{n}}_{x=0}=1$ from which Taylor series for $\displaystyle e^x = \ln^{1}(x)$ follows. But I do have a question (surprise): Show, $\displaystyle \ln ab = \ln a + \ln b$ from the lntegral definition, i.e., Show $\displaystyle \int_{1}^{ab}\frac{dx}{x}= \int_{1}^{a}\frac{dx}{x}+\int_{1}^{b}\frac{dx}{x}$  skipjack has a way of editing typography so that text and equations are uniform. How? Last edited by skipjack; December 11th, 2017 at 01:44 PM. 
December 11th, 2017, 08:30 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,385 Thanks: 2476 Math Focus: Mainly analysis and algebra 
\begin{align*} \int_1^{ab}\frac1t\,\mathrm dt &= \int_1^{a}\frac1t\,\mathrm dt + \int_a^{ab}\frac1t\,\mathrm dt \\ &= \int_1^{a}\frac1t\,\mathrm dt + \int_1^{b}\frac1u\,\mathrm du & \text{(where $u=at$)} \end{align*} 
December 11th, 2017, 09:11 AM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,444 Thanks: 106 
Actually, u=t/a, but it's the thought that counts. 10/10, no deduction for typo. Many thanks.

December 11th, 2017, 12:30 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,385 Thanks: 2476 Math Focus: Mainly analysis and algebra 
Oh, well perhaps some admin can correct it.

December 11th, 2017, 01:14 PM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,444 Thanks: 106 
Exponents from scratch based on above. (till now, e$\displaystyle ^{x}$ is not an exponential, it is just symbol for inverse of $\ln x$ which is defined by integral.) $\displaystyle a^{x} = e^{a\ln x}$, which is a definition in terms of known, calculable, quantities. Otherwise, what is .8375^{4.2319}? Usual laws of exponents should follow. $\displaystyle \ln_{a}a^{x}=x$ is definition of inverse function ln$\displaystyle _{a}$. (Since $\displaystyle \ln e^{x}x, \ln \equiv \ln_{e}$). $\displaystyle a^{\ln_{a}x}=x$, and take ln of both sides. $\displaystyle \ln_{a}x(\ln a)=\ln x$, and $\displaystyle \ln_{10}x=\ln x/\ln10$. Last edited by skipjack; December 11th, 2017 at 01:26 PM. 
December 11th, 2017, 01:49 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,521 Thanks: 1747 
Obviously, $\displaystyle a^{x} = e^{a\ln x}$ is wrong. I assume you meant $\displaystyle a^{x} = e^{x\ln a}\!$.

December 11th, 2017, 03:03 PM  #7  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,444 Thanks: 106  Quote:
Amazingly, if you calculate $\displaystyle 3^{2}=e^{2\ln3}$ on a simple TI calculator, it comes up with 9. Didn't expect that, after all the ten place intermediate decimals. Guess it was just a matter of rounding up the final number to ten places, which was 9.  
December 11th, 2017, 05:00 PM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,444 Thanks: 106 
On the other hand, consider the function $\displaystyle f(x)=x^{a}$ with definition: $\displaystyle f(x)=x^{a}=e^{a\ln x}$ What is $\displaystyle f^{1}(f(x))=x$? $\displaystyle f^{1}(x) = x^{\frac{1}{a}}$ $\displaystyle f^{1}(x^{a})= (x^{a})^{\frac{1}{a}}=(e^{a\ln x})^{\frac{1}{a}}=x$ To summarize, $\displaystyle a^{x}$ and $\displaystyle x^{a}$ are different functions of x with different inverses. 
December 11th, 2017, 05:13 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,385 Thanks: 2476 Math Focus: Mainly analysis and algebra  
December 12th, 2017, 07:23 AM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,444 Thanks: 106 
Two ways of looking at $\displaystyle \ln_{a}x$ 1) Def: $\displaystyle \ln_{a}x$ is the power you have to raise a to to get x: $\displaystyle a^{\ln_{a}x}=x$ $\displaystyle \ln_{a}(a^{x}) = x$, power you have to raise a to to get a$\displaystyle ^{x}$. 2) Def: $\displaystyle \ln_{a}x = f^{1}(a^{x})$ $\displaystyle \ln_{a}(a^{x}) = x, f^{1}(f(x))=x$ $\displaystyle a^{\ln_{a}x}=x, f(f^{1}(x))=x$ Def 1 is what I will remember a week from now because that's what I learned. I like Def 2 as a development from first principles as in this thread. 