# Thread: log and exponential questions

1. ## 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?

2. 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

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

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

$X$ here is any function within the domain of the log

3. 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?

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

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

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

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

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

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

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

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

4. 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?

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

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

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

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

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

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

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

5. [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.

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

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

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

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

$= \frac{{1 - 0}}
{{1 - 0}} = 1 \hfill \\$

6. For the first one, if the lim was as x goes to negative infinity, would it also be one, or negative one?