# Thread: Problem Solving - Log. Related Questions

1. ## Problem Solving - Log. Related Questions

Hey Guyzz,

I Just need help with 2 problem solving question related to logarithms:

1) The average number of customer, c, at a 24-hour sandwich shop per hour is modelled roughly by the equation $-5cos [\frac{(pi)h}{12}] + 12$ , with h = 0 representing midnight. How many hours per day are there at least 13 customer per hour, to the nearest hour? What time of day is peak business?

2) The relation between the number of time in days, D, it takes for a planet to revolve around the Sun is related to the planet's average distance from the Sun, k, in millions of kilometers is defined by $logD = \frac{3}{2}logk - 0.7.$ How much longer does it take Mars than Venus to orbit the Sun if Mars is 207 million km and Venus is 108 million km away from the Sun?

Please guyzz, try your best to solve this problem, I would really appreciate it if you guyz would help me out.
Thank You.

2. Originally Posted by MordernWar2
Hey Guyzz,

I Just need help with 2 problem solving question related to logarithms:

1) The average number of customer, c, at a 24-hour sandwich shop per hour is modelled roughly by the equation $-5cos [\frac{(pi)h}{12}] + 12$ , with h = 0 representing midnight. How many hours per day are there at least 13 customer per hour, to the nearest hour? What time of day is peak business?
Since that formula gives "average" number of customers, it won't really tell you anything about "at least" but I think what you are saying is that you want $-5 cos(\frac{\pi h}{12})+ 12\ge 13$ which is, of course, equivalent to $-5 cos(\frac{\pi h}{12})\ge 1$ which is, again, equivalent to $cos(\frac{\pi h}{12})\le 1/5$.

Now, the simplest way to handle a complicated inequality like that is to first solve the associated equation, $cos(\frac{\pi h}{12})= 1/5$. Those points separate ">" from "<" and you can try a point in each interval to determine which is true for that interval.

(I see nothing to do with "logarithms" in this problem.)

2) The relation between the number of time in days, D, it takes for a planet to revolve around the Sun is related to the planet's average distance from the Sun, k, in millions of kilometers is defined by $logD = \frac{3}{2}logk - 0.7.$ How much longer does it take Mars than Venus to orbit the Sun if Mars is 207 million km and Venus is 108 million km away from the Sun?

Please guyzz, try your best to solve this problem, I would really appreciate it if you guyz would help me out.
Thank You.
How about you try your best? I assume you have tried but it would help us if you would at least show what you have tried!

Replacing "k" by 207000000 gives $log(D)= \frac{2}{3}log(207000000)- 0.7$. You should be able to calculate the right side, then take 10 to that power (I am assuming this is the common logarithm) to find D. Do the same for k= 108000000 and subtract to find "how much longer".

3. Originally Posted by HallsofIvy
Since that formula gives "average" number of customers, it won't really tell you anything about "at least" but I think what you are saying is that you want $-5 cos(\frac{\pi h}{12})+ 12\ge 13$ which is, of course, equivalent to $-5 cos(\frac{\pi h}{12})\ge 1$ which is, again, equivalent to $cos(\frac{\pi h}{12})\le 1/5$.

Now, the simplest way to handle a complicated inequality like that is to first solve the associated equation, $cos(\frac{\pi h}{12})= 1/5$. Those points separate ">" from "<" and you can try a point in each interval to determine which is true for that interval.

(I see nothing to do with "logarithms" in this problem.)

How about you try your best? I assume you have tried but it would help us if you would at least show what you have tried!

Replacing "k" by 207000000 gives [tex]log(D)= \frac{2}{3}log(207000000)- 0.7[tex]. You should be able to calculate the right side, then take 10 to that power (I am assuming this is the common logarithm) to find D. Do the same for k= 108000000 and subtract to find "how much longer".
Hey Thanks for all of this. So far I get #2) and I got the answer to for it.. but the one i dont get it the first problem, Im kinda having problem with it.. Can u please help me out.

4. Well, what did you get as solutions for $cos(\frac{\pi h}{12})= 1/5
$
? What does your calculator say $cos^{-1}(.20)$ is? (Be sure your calculator is in radian mode.)