Solving for n: power and log rules refresh

What would you get?In summary, to solve for n in the equation n log n = x, you can use the approximation that x and n are of similar magnitude. This allows you to substitute one occurrence of n with x, resulting in the simplified equation n = x/log x.
  • #1
nrobidoux
25
0

Homework Statement



A computer can perform 1010 ops/s. Assume 1 op per 1 input. Given the following algorithmic complexities how many inputs can be performed in an hour.

  • n2
  • n3
  • 100n2
  • n log n
  • 2n
  • 22n

Homework Equations

The Attempt at a Solution


[/B]
C = 1·1010 ops/s · 3.6·102 s/hr
C = 3.6·1012 ops/hr

Set each complexity equal to C.

I know this is almost basic algebra but I haven't done this in decades. I think I'm good in the first 3... it's the last 3 I'm a bit confused on. Mostly n log n. Did some googling of the rules but I'm still confused there, and I've never seen a log of a log...

----------------------------------------------
n2 = C
n
(2 · 1/2) = C 1/2
n = C 1/2

----------------------------------------------
The next is basically the same except it's cubic: n = C 1/3

----------------------------------------------
100n2 = C
n
2 = C/100
n(2 · 1/2) = (C/100)1/2
n = (C/100)1/2

----------------------------------------------
n logb( n ) = C

... (more Googling)...


I have no clue... I found another rule. On the surface I think I applied them correctly but I don't know how to get n by itself. I got: bC = nn and bC/n = n... of course if I really looked at those two...

----------------------------------------------
2n = C
n
= log2 ( C )

----------------------------------------------
22n = C
2n = log2 ( C )
n = log2 ( log2 ( C ))
 
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  • #2
nrobidoux said:
3.6·102 s/hr
Check that.
nrobidoux said:
n logb( n ) = C
No base was given.
You only need a solution that is asymptotically correct. Given y log y = x, make a guess as to what y is, approximately, as a function of x.
 
  • #3
nrobidoux said:

Homework Statement



A computer can perform 1010 ops/s. Assume 1 op per 1 input. Given the following algorithmic complexities how many inputs can be performed in an hour.

  • n2
  • n3
  • 100n2
  • n log n
  • 2n
  • 22n

Homework Equations

The Attempt at a Solution


[/B]
C = 1·1010 ops/s · 3.6·102 s/hr
C = 3.6·1012 ops/hr

Set each complexity equal to C.----------------------------------------------
n logb( n ) = C

... (more Googling)...


I have no clue... I found another rule. On the surface I think I applied them correctly but I don't know how to get n by itself. I got: bC = nn and bC/n = n... of course if I really looked at those two...

The solution to ##n \log_2(n) = C## involves the so-called Lambert W-function, which is non-elementary (not a simple power, root, trig, exponential, etc.) Maple gets the solution
$$n = \frac{C \ln(2)}{W(C \ln(2))}, $$
where ##W(w)## is the Lambert-W function of ##w##.
 
  • #4
I used https://www.desmos.com/calculator to get an answer. I'll ask the professor for some clarification.

n = 1.13 ·1011
for n log10 ( n ) = C

The Lambert-W function as part of the solution sounds a bit too complicated for a day 1 homework problem. I think all these were intended to be solved with a pen and paper. Was the application of log rules applied correctly to the last problem? The 22n?

As for the 3.6·102 s/hr ... easy peasy with simple tricks. 60 sec/min x 60 min/hr.. units cancel to sec/hr 6X6 = 36. Add 2 zeroes. 3.6·102 Honestly thought it was 36·102 but Google docs is converting 3.6E2 to ... err well yesterday it was confusing me. So changing the power to 13 from 12 in C. I get 2.9 ·1012 above.
 
  • #5
nrobidoux said:
The Lambert-W function as part of the solution sounds a bit too complicated for a day 1 homework problem
As I posted, you only need an approximation. Go on, make a guess as to how y depends on x, roughly, if y log y = x.
 
  • #6
nrobidoux said:
I used https://www.desmos.com/calculator to get an answer. I'll ask the professor for some clarification.

n = 1.13 ·1011
for n log10 ( n ) = C

The Lambert-W function as part of the solution sounds a bit too complicated for a day 1 homework problem. I think all these were intended to be solved with a pen and paper. Was the application of log rules applied correctly to the last problem? The 22n?

As for the 3.6·102 s/hr ... easy peasy with simple tricks. 60 sec/min x 60 min/hr.. units cancel to sec/hr 6X6 = 36. Add 2 zeroes. 3.6·102 Honestly thought it was 36·102 but Google docs is converting 3.6E2 to ... err well yesterday it was confusing me. So changing the power to 13 from 12 in C. I get 2.9 ·1012 above.

Obviously we are talking about two different problems! I gave you a formula for solution that involved the symbolic C, just as you had done for problems A-C and E-F. I was pointing out to you that a symbolic answer involves a new function that you probably have never heard of before (but which is, in fact, used a lot in combinatorics and computer science). Of course if somebody gives you a numerical value of C you have a completely different situation, where some good guesses and a few trial values will get you a usable solution quickly. But, you need a numerical value for C in order to do that.

If your computer is loaded with a spreadsheet such as EXCEL, you can solve such equations by calling on the "Solver" tool, so you don't need to go to some on-line package if you would rather not.
 
  • #7
I'll see if my version has that. It's an old version of Excel. I can be a bit slow when it comes to math... I have the answer but just want to be able to understand the process to get there in case it pops up on a test. It happens to be proctored online somehow. I asked for some clarification on the base... see what happens. I got quite a few days before this is due.

C = 3600 * 10^10

I appreciate the help. Forum is awesome. :D
 
  • #8
haruspex said:
As I posted, you only need an approximation. Go on, make a guess as to how y depends on x, roughly, if y log y = x.
I'll add a hint. To a first approximation, x and y would be of similar magnitude. Use that to substitute for one of the occurrences of y.
 

Related to Solving for n: power and log rules refresh

1. What are the basic rules for solving for n in exponential equations?

The basic rules for solving for n in exponential equations include using the power rule, logarithm rule, and inverse property of exponents. The power rule states that when raising a power to another power, you can multiply the exponents. The logarithm rule states that when solving for n in an exponential equation, you can take the logarithm of both sides to remove the exponent. The inverse property of exponents states that when an exponential equation has the same base on both sides, you can set the exponents equal to each other to solve for n.

2. How do I use the power rule to solve for n in exponential equations?

To use the power rule, you must have an equation in the form of a^n = b, where a is the base and b is the result of the exponent. You can then take the logarithm of both sides and use the logarithm rule to remove the exponent. This will leave you with the equation loga(b) = n, where n is the value of the exponent.

3. Can I solve for n in an exponential equation without using logarithms?

Yes, you can solve for n without using logarithms by using the inverse property of exponents. This method works when the equation has the same base on both sides. You can set the exponents equal to each other and solve for n.

4. What is the difference between solving for n in exponential equations and logarithmic equations?

Solving for n in exponential equations involves finding the value of the exponent, while solving for n in logarithmic equations involves finding the value of the variable within the logarithm. In exponential equations, the base is raised to a certain power, while in logarithmic equations, the base is the result of the exponent.

5. Are there any other methods for solving for n in exponential equations?

Yes, there are other methods for solving for n in exponential equations, such as using the change of base formula or using the properties of logarithms. These methods may be useful when the equation does not have the same base on both sides or when the exponent is a variable.

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