## f(n) = log n t = 1 seconds

Hi,

I'm doing this algorithm questions and i need to find the largest size n of a problem that can be solved in time t, assuming that the algorithm to solve the problem takes f(n) microseconds.

For example:

f(n) = log n
t = 1 seconds

how do i get the largest size of n in t time??

what i did is assume log is base 10, then:

lg x = y
x= 10^y
since:
f(n) = 1,000,000 microseconds
n = 10^1,000,000

am i right??

you can refer to the Q here: http://problems.datastructures.net/d...ning-times.pdf

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 Quote by nuttynibbles Hi, I'm doing this algorithm questions and i need to find the largest size n of a problem that can be solved in time t, assuming that the algorithm to solve the problem takes f(n) microseconds. For example: f(n) = log n t = 1 seconds how do i get the largest size of n in t time?? what i did is assume log is base 10, then: lg x = y x= 10^y since: f(n) = 1,000,000 microseconds n = 10^1,000,000 am i right?? you can refer to the Q here: http://problems.datastructures.net/d...ning-times.pdf
Well, first things first, you need to read the question a bit more carefully. The opening states clearly:

 Note that log n means the logarithm in base 2 of n.
Other than that, you should be on the right track.
 oh my.. tks if it's base 2, then: lg2 x = y x= 2^y since: f(n) = 1,000,000 microseconds n = 2^1,000,000 but why is the ans 10^3000000??

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## f(n) = log n t = 1 seconds

 Quote by nuttynibbles oh my.. tks if it's base 2, then: lg2 x = y x= 2^y since: f(n) = 1,000,000 microseconds n = 2^1,000,000 but why is the ans 10^3000000??
Do you mean 10300,000 as opposed to 103,000,000? If

$$2^{10^6}=10^x$$

find x.
 sorry i meant 300,000 since: 2^(10^6) = 10^x That means: since x = 10^6, therefore 10^x = 10^(10^6). but how do i get the value 10^300000??

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Gold Member
 Quote by nuttynibbles sorry i meant 300,000 since: 2^(10^6) = 10^x That means: 10^(2^(10^6)) correct. but how do i calculate the value 10^300000?? i'll get error on my calculator
Not... quite. You have to take the log base 10 to isolate x, and then you get 10^6 log_10(2), or about 1000000 * log_10(2). That's the x you're looking for, and it's easy to prove that it's about 300000. Then you get that 2^(10^6) is about equal to 10^(300000).
 hi char.limit, may i know which log rule define this??

Recognitions:
Homework Help
 Quote by nuttynibbles hi char.limit, may i know which log rule define this??
If

$$a^b=c$$

then

$$b=\log_ac$$

So can you apply this to your example?
 sorry but im actually still confuse.. i dont really understand when char.limit said i can isolate x by taking log_10 of x and i get 106 . log_10 (2). i change x->y to avoid confusion. since the rule says: y = log_10 x, then x=10y if we continue from 2106 =10y then y = log_10(2(106)) ???? but how did char.limit get log_10(2).106??
 Hi, i finally understood. the rule that i was looking for is loga(Xy) = y.loga(X) therefore, 2106 =10y y = 106 . log10(2) However, i got another Q. if the Q was asked to use log2, must we give the final ans in log10??

Recognitions:
Homework Help
 Quote by nuttynibbles Hi, i finally understood. the rule that i was looking for is loga(Xy) = y.loga(X) therefore, 2106 =10y y = 106 . log10(2)
Yes, that's right

 Quote by nuttynibbles However, i got another Q. if the Q was asked to use log2, must we give the final ans in log10??
No, it's not necessary. I'd usually leave the answer as it is unless I was specifically asked to convert it into that form.
 hmm okay tks alot peeps

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