Beast of a problem...Log: Solve m = (n/(n+1)) * k^ (-1/n) for N

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lekwauwa44

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A young adult I was tutoring asked me this question:

m = 6.29, k = 0.0882

[FONT=&quot]m = (n/(n+1)) * k^ (-1/n) - solve for N. -

I couldn't get around log(n+1).

Any suggestions?[/FONT]
 
lekwauwa44 said:
A young adult I was tutoring asked me this question:

m = 6.29, k = 0.0882

m = (n/(n+1)) * k^ (-1/n) - solve for N. -

I couldn't get around log(n+1).

Any suggestions?
Click to expand...

Four things:

1) "n" and "N" are not the same thing. Thus, it is either very difficult or very easy to solve for "N", as it does not appear in the equation.

2) If you mean "Solve for 'n'", then it cannot be done. With "n" both in and out of the exponent, you cannot resolve with a lovely algebraic expresssion. This does not mean you can't solve it.

3) "Calculus"? This leads me to #4

4) What is it you are trying to do? Is there more to the problem statement? Perhaps your equation is not where you should have landed.
 
lekwauwa44 said:
A young adult I was tutoring asked me this question:

m = 6.29, k = 0.0882

m = (n/(n+1)) * k^ (-1/n) - solve for N. -

I couldn't get around log(n+1).

Any suggestions?
Click to expand...
Can we see your work so we know where you made any mistakes? I am just curious to see what you mean when you said I couldn't get around log(n+1).
 
tkhunny said:
Four things:

1) "n" and "N" are not the same thing. Thus, it is either very difficult or very easy to solve for "N", as it does not appear in the equation.

2) If you mean "Solve for 'n'", then it cannot be done. With "n" both in and out of the exponent, you cannot resolve with a lovely algebraic expression. This does not mean you can't solve it.

3) "Calculus"? This leads me to #4

4) What is it you are trying to do? Is there more to the problem statement? Perhaps your equation is not where you should have landed.
Click to expand...

Yes, solve for "n". The problem can be (according to the teacher) be solved with Calculus or By using the Log function.
 
Jomo said:
Can we see your work so we know where you made any mistakes? I am just curious to see what you mean when you said I couldn't get around log(n+1).
Click to expand...

When I take the Log() of both sides of the equation, it appears manageable until log(n+1) which I am not sure how to "lovely-ly" break down. I found a way to break it down with a Log() identity found on Wikipedia, but it still breaks it down to the log of a fraction involving (n+1) or (n-1).
 
lekwauwa44 said:
Yes, solve for "n". The problem can be (according to the teacher) be solved with Calculus or By using the Log function.
Click to expand...
Unless there is another condition on the equation for n Calculus won't help either. Is this the whole question?

-Dan
 
lekwauwa44 said:
Yes, solve for "n". The problem can be (according to the teacher) be solved with Calculus or By using the Log function.
Click to expand...

Notice how I did not say, "If we are clever enough, we might be able to do it."
I said, "...it cannot be done." This latter message remains correct.

Teachers can be mistaken. Students can misunderstand and misquote.

Where did the numbers come from in your written example? You have not shown the entire problem.
 
The whole thread is a bit weird. In the first post, the OP said he was tutoring someone. Later, we are told about a teacher who said "it" can be "solved" by calculus (Newton's method perhaps). In any case, it is not clear that we have a complete and exact problem statement (perhaps because the OP has not seen it).
 
lekwauwa44 said:
… can be (according to the teacher) … solved with Calculus or By using the Log function.
Click to expand...
The equation you posted cannot be solved by simply taking logarithms.

Some calculus students learn about the LambertW function (used with equations containing both algebraic and transcendental functions, like the one you posted), but I agree with others' viewpoints here; something is wrong with the given equation and/or instruction because logarithms alone will not solve it.

n = ln(k) / [LambertW(k∙ln(k)/m) - ln(k)]
 
JeffM said:
The whole thread is a bit weird. In the first post, the OP said he was tutoring someone. Later, we are told about a teacher who said "it" can be "solved" by calculus (Newton's method perhaps). In any case, it is not clear that we have a complete and exact problem statement (perhaps because the OP has not seen it).
Click to expand...

So teachers assign homework to kids and parents of kids get tutors for kids to help them with their homework. I am a tutor helping a student with homework a student was assigned by a teacher. That had to have been clear. Again: does anyone know any ways to solve this problem? Really...any help would be great.
 
lekwauwa44 said:
So teachers assign homework to kids and parents of kids get tutors for kids to help them with their homework. I am a tutor helping a student with homework a student was assigned by a teacher. That had to have been clear. Again: does anyone know any ways to solve this problem? Really...any help would be great.
Click to expand...

You were shown a solution (I haven't verified it); here is what Wolfram Alpha does with it. As you will see, they do something similar, using a special function similar to Lambert W, or else using numerical methods to get a numerical answer.

People have asked for context, particularly the exact wording of the problem and what the student is learning. If we were dealing directly with the student, we would have asked directly: Are you studying numerical methods (which can involve calculus), or have you learned about special functions? What a tutor needs is not "any way to solve the problem" but "how the student can use what he is learning to solve the problem". Without that context, there's no a lot we can do.

Have you gone back to the student to find out the information that is needed to determine what kind of solution is expected?
 
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