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

  • Thread starter Michael Santos
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In summary, the conversation involved trying to simplify the expression e^ln (ln (x+h+3)) using the equation e^x * e^h = e^x+h. However, it was determined that this expression cannot be simplified further. The conversation then shifted to finding the difference quotient of ln (x+3), which involves taking the derivative. It was discussed that the derivative of ln (x+3) is 1/x+3 and that this result can be obtained by using the chain rule. The concept of using a Taylor series was also mentioned to help simplify the expression.
  • #1
Michael Santos
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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))
 
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  • #2
## e^{\ln{y}}=y ## . With that equation, you should be able to simplify this thing.
 
  • #3
Charles Link said:
## 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?
 
  • #4
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.
 
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  • #5
Charles Link said:
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
 
  • #7
Charles Link said:
See the last line I added to post 4.
This is related to finding the difference quotient of ln (x+3). The difference quotient
 
  • #8
What happened to the ## h ## ? i.e. You started with ## \ln(x+h+3) ##.
 
  • #9
Charles Link said:
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
 
  • #10
## \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} ## ).
 
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  • #11
Charles Link said:
## \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?
 
  • #12
Michael Santos said:
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.
 
  • #13
Charles Link said:
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
 
  • #14
If you use what I gave you in post 10, it leads immediately to this result.
 
  • #15
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) ##.
 
  • #16
Charles Link said:
## \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} ## ).
Charles Link said:
## \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)?
 
  • #17
Charles Link said:
If you use what I gave you in post 10, it leads immediately to this result.
Was that only using the difference quotient?
 
  • #18
Michael Santos said:
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.
 
  • #19
Charles Link said:
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.
 
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  • #20
Michael Santos said:
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.
 
  • #21
Mark44 said:
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.
 
  • #22
Michael Santos said:
I apologize
I will not commit this offence twice.
Glad to hear it!
 

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