20091104, 23:53  #1 
"Kyle"
Feb 2005
Somewhere near M50..sshh!
1101111110_{2} Posts 
Exponential Inequality
I came across a math puzzle on the Internet tonight and I was wondering what the method of solution would be.
The inequality is as follows: ln x < x^.1 such that x is > 3 Note: Although there are an infinite number of solutions to this problem, the problem is asking for the smallest integer that satifies the equation. My guess was to use infinite series, though I'm not sure how to set it up. My math knowledge is limited to Differential Equations (Ordinary) and Calc 13. (Basic, not Advanced). Any insight would be greatly appreciated. Thanks Last fiddled with by Primeinator on 20091104 at 23:54 Reason: Added the stipulation that x>3 
20091105, 03:30  #2 
"William"
May 2003
New Haven
2×3^{2}×131 Posts 

20091105, 04:23  #3 
Jun 2003
13×19^{2} Posts 
An NR solution could be obtained, no? By Trial and error, looks like the solution lies between 3.4e15 and 3.5e15
Code:
3430631121407800 35.7715206395729718 35.7715206395729709 3430631121407801 35.7715206395729721 35.7715206395729719 3430631121407802 35.7715206395729724 35.7715206395729730 Last fiddled with by axn on 20091105 at 04:34 
20091105, 05:42  #4  
"Kyle"
Feb 2005
Somewhere near M50..sshh!
2×3×149 Posts 
Quote:


20091105, 06:54  #5 
Aug 2006
31×191 Posts 
Pari gets the solution in < 10 milliseconds, even without transforming the problem.
Code:
solve(x=10,1e99,log(x)x^.1) 
20091105, 06:55  #6  
Jun 2003
11125_{8} Posts 
Quote:
NR is your best bet. Though, since it is an integral solution we are searching for, we can do a simple "halving the interval" technique also. NR requires that you have a reasonable starting position. I proceeded by starting at x=3 and kept on doubling x until I got an upper bound. Then it's just a matter of applying whichever technique you want. Here, with a reasonable starting point, NR will converge in < 5 iterations. PS: This particular problem is relatively easy to solve since both functions are monotonically increasing. 

20091105, 06:56  #7 
Jun 2003
1255_{16} Posts 

20091105, 07:03  #8 
Aug 2006
31×191 Posts 
Brent's method, I think. Secant would be just as good in this case.
NewtonRaphson isn't fast unless you can compute derivatives quickly. The bisection method is slow, especially since you don't know a priori where to search. Last fiddled with by CRGreathouse on 20091105 at 07:05 
20091105, 08:34  #9  
"Richard B. Woods"
Aug 2002
Wisconsin USA
1E0C_{16} Posts 
Quote:


20091105, 10:14  #10  
Jun 2003
1001001010101_{2} Posts 
Quote:
Of course, I only skimmed thru that page. If some one could post the canned solutions of a onevariable quartic, only in terms of its coeffs, much appreciated. EDIT: Or not. Apparently, cubic has closed form solution Last fiddled with by axn on 20091105 at 10:22 

20091105, 13:13  #11 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
axn, you're thinking of degree 5 polynomials (and higher)

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