# log and exponential questions

• Sep 8th 2008, 07:15 PM
3deltat
log and exponential questions
These aren't too hard... they are just some problems that Im not quite sure on. If you can help for each I really would appreciate it..

The first one is...

Find the limit as x goes to infinity for ( (e^3x)-(e^-3x) ) / ( (e^3x)+(e^-3x) )

now i know you are supposed to divide num and denom by e^3x to =
lim x -> oo ( 1-(3^-6x) ) / ( 1+(e^-6x) )
which is basically one minus infinity over 1 plus infinity..

The answer of course, is -1... how do i reason that? Im confused there...

Second question is, Differentiate Log10(x/(x-1))
The first thing i did to make myself confused is thinking that log base 10 x is = lnx.. What is the real relationship? I forgot... and the second thing, could I change the function to f(x) = log base 10 x - log base 10 (x-1) and then differentiate from there?

And the last question is to evaluate the integral of Log base 10 x over x dx
(sorry for having to say it in words)
For this one, the book says something about changing it to lnx/ln10/x... i think... do you have to do it this way?

• Sep 8th 2008, 07:24 PM
Jhevon
Quote:

Originally Posted by 3deltat
These aren't too hard... they are just some problems that Im not quite sure on. If you can help for each I really would appreciate it..

The first one is...

Find the limit as x goes to infinity for ( (e^3x)-(e^-3x) ) / ( (e^3x)+(e^-3x) )

now i know you are supposed to divide num and denom by e^3x to =
lim x -> oo ( 1-(3^-6x) ) / ( 1+(e^-6x) )
which is basically one minus infinity over 1 plus infinity..

The answer of course, is -1... how do i reason that? Im confused there...

if -1 is the answer, you have your signs mixed up somewhere. this limit is 1

Quote:

Second question is, Differentiate Log10(x/(x-1))
The first thing i did to make myself confused is thinking that log base 10 x is = lnx.. What is the real relationship? I forgot... and the second thing, could I change the function to f(x) = log base 10 x - log base 10 (x-1) and then differentiate from there?

And the last question is to evaluate the integral of Log base 10 x over x dx
(sorry for having to say it in words)
For this one, the book says something about changing it to lnx/ln10/x... i think... do you have to do it this way?

for these two, use the change of base formula.

recall, $\displaystyle \log_a b = \frac {\log_c b}{\log_c a}$

in particular, $\displaystyle \log_{10} X = \frac {\ln X}{\ln 10}$

$\displaystyle X$ here is any function within the domain of the log
• Sep 8th 2008, 07:46 PM
Shyam
Second question is, Differentiate Log10(x/(x-1))
The first thing i did to make myself confused is thinking that log base 10 x is = lnx.. What is the real relationship? I forgot... and the second thing, could I change the function to f(x) = log base 10 x - log base 10 (x-1) and then differentiate from there?

$\displaystyle {\text{We know that, }}\boxed {\log _{10} x = \frac{{\ln x}} {{\ln 10}}} \hfill \\$

$\displaystyle f\left( x \right) = \log _{10} \left( {\frac{x} {{x - 1}}} \right) = \log _{10} x - \log _{10} \left( {x - 1} \right) \hfill \\$

$\displaystyle f\left( x \right) = \frac{{\ln x}} {{\ln 10}} - \frac{{\ln \left( {x - 1} \right)}} {{\ln 10}} \hfill \\$

$\displaystyle {\text{Now differentiate,}} \hfill \\$

$\displaystyle f'\left( x \right) = \frac{1} {{x\ln 10}} - \frac{1} {{\left( {x - 1} \right)\ln 10}} \hfill \\$

$\displaystyle = \frac{1} {{\ln 10}}\left( {\frac{1} {x} - \frac{1} {{x - 1}}} \right) \hfill \\$

$\displaystyle = \frac{1} {{\ln 10}}\left( {\frac{{x - 1 - x}} {{x\left( {x - 1} \right)}}} \right) \hfill \\$

$\displaystyle = \frac{{ - 1}} {{x\left( {x - 1} \right)\ln 10}} \hfill \\$
• Sep 8th 2008, 08:02 PM
Shyam
And the last question is to evaluate the integral of Log base 10 x over x dx
(sorry for having to say it in words)
For this one, the book says something about changing it to lnx/ln10/x... i think... do you have to do it this way?

$\displaystyle \int {\frac{{\log _{10} x}} {x}dx} \hfill \\$

$\displaystyle = \int {\frac{{\ln x}} {{x\ln 10}}dx} \hfill \\$

$\displaystyle = \frac{1} {{\ln 10}}\int {\frac{{\ln x}} {x}} dx \hfill \\$

$\displaystyle {\text{Now, let, }}\ln x = y \Rightarrow \frac{1} {x}dx = dy \hfill \\$

$\displaystyle = \frac{1} {{\ln 10}}\int {y\;dy} \hfill \\$

$\displaystyle = \frac{1} {{\ln 10}}\left( {\frac{{y^2 }} {2}} \right) + c \hfill \\$

$\displaystyle = \frac{{\left( {\ln x} \right)^2 }} {{2\ln 10}} + c \hfill \\$
• Sep 8th 2008, 08:15 PM
Shyam
[quote=3deltat;184167]These aren't too hard... they are just some problems that Im not quite sure on. If you can help for each I really would appreciate it..

The first one is...

Find the limit as x goes to infinity for ( (e^3x)-(e^-3x) ) / ( (e^3x)+(e^-3x) )

now i know you are supposed to divide num and denom by e^3x to =
lim x -> oo ( 1-(3^-6x) ) / ( 1+(e^-6x) )
which is basically one minus infinity over 1 plus infinity..

The answer of course, is -1... how do i reason that? Im confused there...

The answer is 1. it is not -1.

$\displaystyle = \mathop {\lim }\limits_{x \to \infty } \frac{{e^{3x} - e^{ - 3x} }} {{e^{3x} + e^{ - 3x} }} \hfill \\$

$\displaystyle {\text{Divide numerator and denominator with }}e^{3x} , \hfill \\$

$\displaystyle = \mathop {\lim }\limits_{x \to \infty } \frac{{1 - e^{ - 6x} }} {{1 + e^{ - 6x} }} \hfill \\$

$\displaystyle = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{1 - \frac{1} {{e^{6x} }}}} {{1 + \frac{1} {{e^{6x} }}}}} \right) \hfill \\$

$\displaystyle = \frac{{1 - 0}} {{1 - 0}} = 1 \hfill \\$
• Sep 9th 2008, 06:55 PM
3deltat
For the first one, if the lim was as x goes to negative infinity, would it also be one, or negative one?