How is e^ln (ln (x+h+3)) broken down

[Michael Santos]

Homework Statement

e^ln (ln (x+h+3))

Homework Equations

For example: e^x * e^h = e^x+h

The Attempt at a Solution

e^ln (ln (x)) * e^ln (ln (h)) * e^ln (ln*3)) does not equal e^ln (ln (x+h+3))

 

[Charles Link]

$e^{\ln{y}}=y$ . With that equation, you should be able to simplify this thing.

 

[Michael Santos]

$e^{\ln{y}}=y$ . With that equation, you should be able to simplify this thing.

Yeah, how about x^2 × x^3 = x^5

What about the original equation e^ln (ln (x+h+3))
How did the inner-most function combine?

 

[Charles Link]

You can't do anything with that. As a homework helper, I am not allowed to give you the answer, but $\ln(a+b+c)=\ln(a+b+c)$. There is nothing you can do to factor that part. $\\$ $\ln(ab)=\ln(a)+\ln(b)$, but that doesn't help you here. $\\$ If you give me what you think is the answer, I will be happy to verify it.

 

[Michael Santos]

You can't do anything with that. As a homework helper, I am not allowed to give you the answer, but $\ln(a+b+c)=\ln(a+b+c)$. There is nothing you can do to factor that part. $\\$ $\ln(ab)=\ln(a)+\ln(b)$, but that doesn't help you here.

Ok, so the function of a natural log as a power can not be changed or modified

 

[Charles Link]

See the last line I added to post 4.

 

[Michael Santos]

See the last line I added to post 4.

This is related to finding the difference quotient of ln (x+3). The difference quotient

 

[Charles Link]

What happened to the $h$ ? i.e. You started with $\ln(x+h+3)$.

 

[Michael Santos]

What happened to the $h$ ? i.e. You started with $\ln(x+h+3)$.

Okay

d/dx[ln (x+3)] =

lim h-->0 of (ln(x+h+3) - ln (x+3))/h

This can me modified to

lim h-->0 of (e^ln (ln (x+h+3)) - e^ln (ln (x+3)))/h
Because e^lny = y

Or you can keep going with the expression that had not been modified

 

[Charles Link]

$\ln(1+h) \approx h$ for small $h$. Here, it helps to have Taylor series concepts. Otherwise, it is somewhat difficult to know exactly how basic you want to keep the operations to solve the problem at hand. $\\$ e.g. Then you can say $\ln(x+3+h)=\ln(x+3) +\ln(1+\frac{h}{x+3}) \approx \ln(x+3)+\frac{h}{x+3}$. (In the first step, I'm using $\ln(ab)=\ln{a}+\ln{b}$ ).

 

[Michael Santos]

$\ln(1+h) \approx h$ for small $h$. Here, it helps to have Taylor series concepts. Otherwise, it is somewhat difficult to know exactly how basic you want to keep the operations to solve the problem at hand.

So it is impossible to find the derivative of ln (x+3) with the difference quotient?

 

[Charles Link]

So it is impossible to find the derivative of ln (x+3) with the difference quotient?

Please define the difference quotient. See also what I added above. You are simply taking a derivative. You can take $y=e^{\ln(y)}$ as a mathematical step if you want. It is perfectly legitimate, but it may be extra work for this problem.

 

[Michael Santos]

Please define the difference quotient. See also what I added above. You are simply taking a derivative. You can take $y=e^{\ln(y)}$ as a mathematical step if you want. It is perfectly legitimate, but may be extra work for this problem.

The definition of the difference quotient is (f(x+h) - f (x))/h where f(x)= ln (x+3)
So
(ln(x+h+3) - ln(x+3))/h
This can be simplified even further but it does not lead to the answer i want which is
1/x+3

 

[Charles Link]

If you use what I gave you in post 10, it leads immediately to this result.

 

[Charles Link]

Otherwise, I'm not sure what you are looking for. Given $y=e^x$, and $\frac{dy}{dx}=e^x$, the result is $x=\ln{y}$ , so that $\frac{dx}{dy}=\frac{1}{e^x}=\frac{1}{y}$, so that $\frac{d \ln{x}}{dx}=\frac{1}{x}$. This stuff is all very consistent. From there you can use the chain rule in working with $y=\ln(x+3)$.

 

[Michael Santos]

$\ln(1+h) \approx h$ for small $h$. Here, it helps to have Taylor series concepts. Otherwise, it is somewhat difficult to know exactly how basic you want to keep the operations to solve the problem at hand. $\\$ e.g. Then you can say $\ln(x+3+h)=\ln(x+3) +\ln(1+\frac{h}{x+3}) \approx \ln(x+3)+\frac{h}{x+3}$. (In the first step, I'm using $\ln(ab)=\ln{a}+\ln{b}$ ).

$\ln(1+h) \approx h$ for small $h$. Here, it helps to have Taylor series concepts. Otherwise, it is somewhat difficult to know exactly how basic you want to keep the operations to solve the problem at hand. $\\$ e.g. Then you can say $\ln(x+3+h)=\ln(x+3) +\ln(1+\frac{h}{x+3}) \approx \ln(x+3)+\frac{h}{x+3}$. (In the first step, I'm using $\ln(ab)=\ln{a}+\ln{b}$ ).

? Are you using other methods to solve this? Can this be solved only using the difference quotient?

And how does ln (x+h+3)= ln(x+3) + ln (1+h/x+3)?

 

[Michael Santos]

If you use what I gave you in post 10, it leads immediately to this result.

Was that only using the difference quotient?

 

[Charles Link]

Was that only using the difference quotient?

Whether you use $e^h \approx 1+h$, or $\ln(1+h) \approx h$, you do need something in addition to the definition of the derivative to solve this. I can't quite tell from the context what kind of additional operations you are allowing. These two approximate results can be readily proven, but the simplest way to prove them is with Taylor series.

 

[Michael Santos]

Whether you use $e^h \approx 1+h$, or $\ln(1+h) \approx h$, you do need something in addition to the definition of the derivative to solve this. I can't quite tell from the context what kind of additional operations you are allowing. These two approximate results can be readily proven, but the simplest way to prove them is with Taylor series.

Thank you charles.
Ok charles i have figured it out
I had to factor the x+3 expression and then use properties of logs the rest was easy to do thank you for your help.

 

[Mark44]

This is related to finding the difference quotient of ln (x+3). The difference quotient

In which case, you should not start a new thread. The expression you started this thread with appears to be due to an error on your part, and doesn't seem to have anything to do with finding the derivative of $\ln(x + 3)$. If you hadn't started a new thread, that could have been straightened out much more quickly.

 

[Michael Santos]

In which case, you should not start a new thread. The expression you started this thread with appears to be due to an error on your part, and doesn't seem to have anything to do with finding the derivative of $\ln(x + 3)$. If you hadn't started a new thread, that could have been straightened out much more quickly.

I apologize
I will not commit this offence twice.

 

[Mark44]

I apologize
I will not commit this offence twice.

Glad to hear it!