Natural log

**matty888** · Apr 26th 2008, 09:11 AM

Define the natural logarithm of a number x in terms of an integral.Show that
ln(xy) = ln(x) + ln(y)
for all x,y >0

**Mathstud28** · Apr 26th 2008, 09:17 AM

Originally Posted by matty888

Define the natural logarithm of a number x in terms of an integral.Show that
ln(xy) = ln(x) + ln(y)
for all x,y >0

$\ln(x)\equiv\int_1^{x}\frac{1}{y}dy$

**TheEmptySet** · Apr 27th 2008, 08:49 AM

Originally Posted by matty888

Define the natural logarithm of a number x in terms of an integral.Show that
ln(xy) = ln(x) + ln(y)
for all x,y >0

define $\ln(x)=\int_{1}^{x}\frac{1}{t}dt$

we want to show $\ln(xy)=\ln(x)+\ln(y)$

so we start with

$\ln(xy)=\int_{1}^{xy}\frac{1}{t}dt$

using the linear property of integrals we get

$\int_{1}^{xy}\frac{1}{t}dt=\int_{1}^{x}\frac{1}{t} dt+\int_{x}^{xy}\frac{1}{t}dt$

lets focus on the 2nd integral

$\int_{x}^{xy}\frac{1}{t}dt$

let $u=\frac{t}{x} \iff ux =t$

$du=\frac{1}{x}dt \iff xdu=dt$

$\int_{x}^{xy}\frac{1}{t}dt=\int_{1}^{y}\frac{1}{ux }xdu=\int_{1}^{y}\frac{1}{u}du=\ln(y)$

so finally with all the above work

$\ln(xy)=\int_{1}^{xy}\frac{1}{t}dt=\int_{1}^{x}\fr ac{1}{t}dt+\int_{1}^{y}\frac{1}{u}du=\ln(x)+\ln(y)$

Yeah!!

**matty888** · Apr 27th 2008, 09:34 AM

Thank you very much TheEmptySet,you made it very clear and understandable,if you could help me with some of my other post i would be very grateful

**Mathstud28** · Apr 30th 2008, 05:55 PM

Originally Posted by matty888

Define the natural logarithm of a number x in terms of an integral.Show that
ln(xy) = ln(x) + ln(y)
for all x,y >0

Isnt this legal for the proof?

$\ln(xy)=\ln\bigg(e^{\ln(x)}e^{\ln(y)}\bigg)=\ln\bi gg(e^{\ln(x)+\ln(y)}\bigg)=ln(x)+\ln(y)$

This just popped into my head on the way home? Its so easy...but is it mathematically legal?

**Jhevon** · Apr 30th 2008, 06:05 PM

Originally Posted by Mathstud28

Isnt this legal for the proof?

$\ln(xy)=\ln\bigg(e^{\ln(x)}e^{\ln(y)}\bigg)=\ln\bi gg(e^{\ln(x)+\ln(y)}\bigg)=ln(x)+\ln(y)$

This just popped into my head on the way home? Its so easy...but is it mathematically legal?

yes it is

**Mathstud28** · Apr 30th 2008, 06:12 PM

Originally Posted by Jhevon

yes it is

Haha...sweet..I thought it was easy

**ThePerfectHacker** · Apr 30th 2008, 06:21 PM

Originally Posted by Mathstud28

Its so easy...but is it mathematically legal?

No it is not. How is $e^x$ defined? In fact, $e^x$ is defined to be the inverse function of $\ln x$ but you never demonstrated it.
---
Here is how to prove it. Let $y>0$ . Define a function $f<img src=$ 0,\infty)\mapsto \mathbb{R}" alt="f

0,\infty)\mapsto \mathbb{R}" /> as $f(x) = \ln (xy) - \ln x$ . This function is differenciable and furthermore, $f'(x) = \frac{y}{yx} - \frac{1}{x} = 0$ . Thus, $f(x) = k$ for some number $k$ . In particular, $f(1) = k$ but $f(1) = \ln (y) - \ln (1) = \ln y$ . Thus, $k=\ln y$ . Therefore, $\ln (xy) - \ln x = \ln y$ .

**CaptainBlack** · May 1st 2008, 01:31 PM

Originally Posted by Mathstud28

Isnt this legal for the proof?

$\ln(xy)=\ln\bigg(e^{\ln(x)}e^{\ln(y)}\bigg)=\ln\bi gg(e^{\ln(x)+\ln(y)}\bigg)=ln(x)+\ln(y)$

This just popped into my head on the way home? Its so easy...but is it mathematically legal?

No because you have to use the definition in terms of the integral, you do not know any other definition for the purposes of this question. That is you need a proof like that of TheEmptySet's post.

RonL

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thank you!!

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