- #1
osqen
- 6
- 0
Homework Statement
e^(0,001*a)-((R^2)*(e^(-0,001*a)))
R and a is constant
i can't take ln of this function, pls help
Solving e^(0.001a) - (R^2)*(e^(-0.001a)): Help Needed!
- Thread starter osqen
- Start date
Click For Summary
Homework Help Overview
The discussion revolves around the expression e^(0.001a) - (R^2)*(e^(-0.001a)), where R and a are constants. Participants are exploring the challenges of taking the natural logarithm of this expression.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Participants discuss the difficulty of taking the natural logarithm of the given expression and question the nature of the problem. Some suggest rewriting parts of the expression to facilitate understanding, while others seek clarification on the original poster's specific question.
Discussion Status
The discussion is ongoing, with various interpretations of the problem being explored. Some participants have offered guidance on rewriting the expression, but there is no clear consensus on the original poster's confusion or the next steps to take.
Contextual Notes
There appears to be some ambiguity regarding the original poster's question and the specific difficulties they are facing with the expression. Additionally, the nature of the constants R and a is not fully defined in the context of the discussion.
e^(0,001*a)-((R^2)*(e^(-0,001*a)))
R and a is constant
i can't take ln of this function, pls help
- #2
nicksauce
Science Advisor
Homework Helper
- 1,270
- 7
you can write R^2e^(-0.001a) as e^(ln(R^2)*-0.001a)
can you take it from there?
can you take it from there?
- #3
sutupidmath
- 1,629
- 4
osqen said:
What exactly is your question? I don't see anythig being asked here.
- #4
osqen
- 6
- 0
my question is x = e^(0,001*a)-((R^2)*(e^(-0,001*a)))
and lnx = ?
and lnx = ?
- #5
quantumdude
Staff Emeritus
Science Advisor
Gold Member
- 5,564
- 24
I still don't understand what the problem here is. If [itex]x=\exp(0.001a)-R^2\exp(-0.001a)[/itex], then obviously:
[itex]\ln(x)=\ln(\exp(0.001a)-R^2\exp(-0.001a))[/itex].
Am I missing something?
[itex]\ln(x)=\ln(\exp(0.001a)-R^2\exp(-0.001a))[/itex].
Am I missing something?
- #6
osqen
- 6
- 0
i lost my ability :( thank you:)
- #7
HallsofIvy
Science Advisor
Homework Helper
- 42,895
- 983
If you were wondering if it could be simplified at all, the answer is "no".
- #8
osqen
- 6
- 0
is it the eqn. as simple as it can be
Similar threads
- · Replies 3 ·
- · Replies 6 ·
- · Replies 7 ·