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 December 11th, 2017, 09:18 AM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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 02:44 PM.
 December 11th, 2017, 09:30 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 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*} Thanks from Maschke and zylo
 December 11th, 2017, 10:11 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Actually, u=t/a, but it's the thought that counts. 10/10, no deduction for typo. Many thanks.
 December 11th, 2017, 01:30 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra Oh, well perhaps some admin can correct it.
 December 11th, 2017, 02:14 PM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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 02:26 PM.
 December 11th, 2017, 02:49 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2337 Obviously, $\displaystyle a^{x} = e^{a\ln x}$ is wrong. I assume you meant $\displaystyle a^{x} = e^{x\ln a}\!$. Thanks from zylo
December 11th, 2017, 04:03 PM   #7
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Quote:
 Originally Posted by skipjack Obviously, $\displaystyle a^{x} = e^{a\ln x}$ is wrong. I assume you meant $\displaystyle a^{x} = e^{x\ln a}\!$.
Indeed. Thanks for catching that.

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, 06:00 PM #8 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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, 06:13 PM   #9
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Quote:
 Originally Posted by zylo To summarize, $\displaystyle a^{x}$ and $\displaystyle x^{a}$ are different functions of x with different inverses.
Pure genius. Who'd've thought it?

 December 12th, 2017, 08:23 AM #10 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 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.

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