- #1
Michael Santos
- 29
- 3
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))
Yeah, how about x^2 × x^3 = x^5Charles Link said:## e^{\ln{y}}=y ## . With that equation, you should be able to simplify this thing.
Ok, so the function of a natural log as a power can not be changed or modifiedCharles 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.
This is related to finding the difference quotient of ln (x+3). The difference quotientCharles Link said:See the last line I added to post 4.
OkayCharles Link said:What happened to the ## h ## ? i.e. You started with ## \ln(x+h+3) ##.
So it is impossible to find the derivative of ln (x+3) with the difference quotient?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.
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 said:So it is impossible to find the derivative of ln (x+3) with the difference quotient?
The definition of the difference quotient is (f(x+h) - f (x))/h where f(x)= ln (x+3)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.
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?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} ## ).
Was that only using the difference quotient?Charles Link said:If you use what I gave you in post 10, it leads immediately to this result.
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 said:Was that only using the difference quotient?
Thank you charles.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.
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 said:This is related to finding the difference quotient of ln (x+3). The difference quotient
I apologizeMark44 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.
Glad to hear it!Michael Santos said:I apologize
I will not commit this offence twice.
The purpose of breaking down e^ln (ln (x+h+3)) is to simplify the expression and make it easier to solve for the value of x. By breaking it down, we can eliminate any unnecessary operations and focus on the key components of the expression.
To break down e^ln (ln (x+h+3)), we can use the properties of logarithms and exponents. First, we can rewrite e^ln (ln (x+h+3)) as e^(ln(ln(x) + ln(h+3))). Then, we can further break down the expression by using the property ln(a+b) = ln(a) + ln(b) to simplify it to e^(ln(ln(x)) + ln(ln(h+3))).
Simplifying expressions like e^ln (ln (x+h+3)) is important because it allows us to better understand the relationship between different mathematical operations and make solving equations easier. It also helps us to avoid mistakes and reduces the chances of getting incorrect results.
For example, let's say we have e^ln (ln (5+2+3)). We can break this down by first rewriting it as e^(ln(ln(5) + ln(2+3))). Then, using the property ln(a+b) = ln(a) + ln(b), we can simplify it to e^(ln(ln(5)) + ln(ln(2+3))). Finally, we can use a calculator to evaluate ln(5) and ln(2+3), and then take the natural exponent of their sum to get the final result.
Breaking down e^ln (ln (x+h+3)) has several benefits, including simplifying the expression, making it easier to solve for the value of x, and better understanding the underlying mathematical principles involved. It also helps to avoid mistakes and allows for more accurate and efficient calculations.