Help With Logarithmic Differentiation

[Civilization]

Hi guys, first post here.

I have a problem that requires me to do logarithmic differentiation, which I haven't learned yet. I tried "Googling" that topic, but I always get confused when reading those websites.

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Let x = the number of workers in a team and T(x) = the time required for the team to finish the job. Since the expression for T(x) will have x occurring in an exponent, use logarithmic differentiation in finding T '(x).

T(x) = ((30)(1.06)[sup:361xyoaa]x-1[/sup:361xyoaa]) / x

This was basically a "minimum" problem where I had to find the number of workers that should be placed on a team so that their job is done in the possible time by the team. What is this least time?. The bolded part is directly quoted from the question. There was also a "suggestion" tidbit that said Work done equals rate of doing work, expressed as work done per hour, times the number of hours worked.

Each worker can do a task in 30 hours. Alone, each worker works at the rate of completing 1/30th of the job per hour, but with other workers, this rate lessens (due to interference). With 2 workers, each worker needs (30)(1.06) hours to get the job done. With 3, each needs (30)(1.06)[sup:361xyoaa]2[/sup:361xyoaa]. The pattern continues.

Here's my first line of work:

t ' (x) = (x-1 * ln (31.8)) / ln x

Do things look good here? And I assume I must set this derivative to equal zero sometime later.

(In case you're wondering, I think I solved this problem by getting x=17. However, I used a brute force method (lol) and would rather use logarithmic differentiation.)

 

[Deleted member 4993]

Hi guys, first post here.

I have a problem that requires me to do logarithmic differentiation, which I haven't learned yet. I tried "Googling" that topic, but I always get confused when reading those websites.

-------------

Let x = the number of workers in a team and T(x) = the time required for the team to finish the job. Since the expression for T(x) will have x occurring in an exponent, use logarithmic differentiation in finding T '(x).

T(x) = ((30)(1.06)[sup:xfv43map]x-1[/sup:xfv43map]) / x

This was basically a "minimum" problem where I had to find the number of workers that should be placed on a team so that their job is done in the possible time by the team. What is this least time?. The bolded part is directly quoted from the question. There was also a "suggestion" tidbit that said Work done equals rate of doing work, expressed as work done per hour, times the number of hours worked.

Each worker can do a task in 30 hours. Alone, each worker works at the rate of completing 1/30th of the job per hour, but with other workers, this rate lessens (due to interference). With 2 workers, each worker needs (30)(1.06) hours to get the job done. With 3, each needs (30)(1.06)[sup:xfv43map]2[/sup:xfv43map]. The pattern continues.

Here's my first line of work:

t ' (x) = (x-1 * ln (31.8)) / ln x

Do things look good here? And I assume I must set this derivative to equal zero sometime later.

(In case you're wondering, I think I solved this problem by getting x=17. However, I used a brute force method (lol) and would rather use logarithmic differentiation.)

If you were not taught logarithmic differentiation - probably you were supposed to use brute force. Anyway:

T(x) = ((30)(1.06)[sup:xfv43map]x-1[/sup:xfv43map]) / x

\(\displaystyle T(x) \, = \, \frac{30\cdot (1.06)^{x-1}}{x}\)

\(\displaystyle ln(T) \, = ln(30) + (x-1)\cdot ln(1.06) - ln(x)\)

\(\displaystyle T' = \frac{30\cdot (1.06)^{x-1}}{x} \cdot (ln(1.06) - \frac{1}{x})\)

so for T' = 0

\(\displaystyle ln(1.06) - \frac{1}{x} = 0\)

Now continue....

By the way x = 17 is a good number

 

[Civilization]

Hi guys, first post here.

I have a problem that requires me to do logarithmic differentiation, which I haven't learned yet. I tried "Googling" that topic, but I always get confused when reading those websites.

-------------

Let x = the number of workers in a team and T(x) = the time required for the team to finish the job. Since the expression for T(x) will have x occurring in an exponent, use logarithmic differentiation in finding T '(x).

T(x) = ((30)(1.06)[sup:2c225c9f]x-1[/sup:2c225c9f]) / x

This was basically a "minimum" problem where I had to find the number of workers that should be placed on a team so that their job is done in the possible time by the team. What is this least time?. The bolded part is directly quoted from the question. There was also a "suggestion" tidbit that said Work done equals rate of doing work, expressed as work done per hour, times the number of hours worked.

Each worker can do a task in 30 hours. Alone, each worker works at the rate of completing 1/30th of the job per hour, but with other workers, this rate lessens (due to interference). With 2 workers, each worker needs (30)(1.06) hours to get the job done. With 3, each needs (30)(1.06)[sup:2c225c9f]2[/sup:2c225c9f]. The pattern continues.

Here's my first line of work:

t ' (x) = (x-1 * ln (31.8)) / ln x

Do things look good here? And I assume I must set this derivative to equal zero sometime later.

(In case you're wondering, I think I solved this problem by getting x=17. However, I used a brute force method (lol) and would rather use logarithmic differentiation.)

If you were not taught logarithmic differentiation - probably you were supposed to use brute force. Anyway:

T(x) = ((30)(1.06)[sup:2c225c9f]x-1[/sup:2c225c9f]) / x

\(\displaystyle T(x) \, = \, \frac{30\cdot (1.06)^{x-1}}{x}\)

\(\displaystyle ln(T) \, = ln(30) + (x-1)\cdot ln(1.06) - ln(x)\)

\(\displaystyle T' = \frac{30\cdot (1.06)^{x-1}}{x} \cdot (ln(1.06) - \frac{1}{x})\)

so for T' = 0

\(\displaystyle ln(1.06) - \frac{1}{x} = 0\)

Now continue....

By the way x = 17 is a good number

Possibly ... but logarithmic diff. would be nice to know in advance, I guess.


\(\displaystyle ln(1.06) = \frac{1}{x}\)

\(\displaystyle x = \frac{1}{ln(1.06)}\)

\(\displaystyle x = 17.1618 ... or x=17 since I have to round down for the number of workers.\)

Thanks Subhotosh Khan. I understand now.

 

[Civilization]

Wait hang on ... I have a question. How did you get from step 2 to step 3 where you have T' = 0?

 

[Deleted member 4993]

Wait hang on ... I have a question. How did you get from step 2 to step 3

That's where logarithmic differentiation happens!

where you have T' = 0?

 

[Civilization]

But you have ln(T) and then it turns into T'

 

[Civilization]

Wait ... I think I understand how you got ln1.06 - 1/x, but I'm unsure as how you got the first part since it's just a copy of the original equation, not the derivative.