OT Weekend Challenge  Printable Version + HP Forums (https://archived.hpcalc.org/museumforum) + Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum1.html) + Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum2.html) + Thread: OT Weekend Challenge (/thread147631.html) 
OT Weekend Challenge  Chuck  02272009 Test your algebra skills on this one. Given f(x)=b^x and g(x)=log(b,x) find all possible values for b so that f and g are tangent. The HP's may be able to numerically find b, but it's more challenging on your own. Have a good weekend. :)
CHUCK
Re: OT Weekend Challenge  Allen  02272009 Quote:
Do you mean the Logarithm is tangent at some particular point, at least one point, or is there some other missing piece of information? Unless I'm misread something, it seems the problem statement is incomplete since f and g are functions of x but we are "finding b such that..."
Re: OT Weekend Challenge  Chuck  02272009 The challenge is to find the unique base b which makes the exponential function with that base b, and the logarithm function with that base tangent at a particular point. Finding the point the two functions are tangent at is most likely a necessity.
C
Re: OT Weekend Challenge  Namir  02272009 This is not some homework you need help with? :)
Namir
Re: OT Weekend Challenge  C.Ret  02282009 Quote: It's really like it is the case.
May I suggest to look at e and 1/e.
Re: OT Weekend Challenge  Chuck  02282009 Heavens no. This is a problem I came across from a math archive two years ago. The "answer" was given, and it turned out to be an interesting problem in that I came up with two solutions instead of their one. I'm not teaching calculus this year so I think it would be a little too challenging for my algebra students. Cheers.
Re: OT Weekend Challenge  Chuck  02282009 They have something to with it, but those are not the answers. It's not a really challenging problem, but it does have an interesting solution.
Re: OT Weekend Challenge  Gerson W. Barbosa  02282009 I thought of trying to solve the following system: ln(b)*b^x = 1/(x*ln(b)) ; the first derivatives of f(x) and g(x) must be the same at that point so they are tangent to each other b^x = ln(x)/ln(b) ; both functions must have the same value at the tangent point By replacing b^x with ln(x)/ln(b) in the first equation we get b = e^(1/(x*ln(x))). Well, at this point I got lost as solving for a particular x was not enough to find b such b and x were solutions to the system. Perhaps you might explain why deriving e^(1/(x*ln(x))) and equating it to zero gives x=1/e and allows us to compute b=e^(e), which are solutions to the system.
Gerson.
Re: OT Weekend Challenge  Namir  02282009 Gerson, I followed much the same steps in working with equating the functions and their derivatives. I look at them as two nonlinear equations with two variables. Anyone can solve the system of 2 nonlinear equations?
Namir
Re: OT Weekend Challenge  C.Ret  03022009 Shall I propose to start over from start: Given f(x) = b^{x} and g(x) = log_{b}(x), we have to found all possible value of b so that given function f and g are tangent. Domain of definition: As a first approach, I suggest to only consider the real space of definition of f and g. The complex space may be investigating the same way and may lead to supplementary solutions, but a nonobvious graphical approach have to be explained. The (real) natural logarithms ln x is only define for strictly positive real argument x. Since log_{b} x = ln(x)/ln(b), g is real only for b>0 and x>0 and undefined (division by zero) at b=1. This greatly simplifies the domain of definition for f as a real function too. Since b^{x} is real for any (strictly) positive value of b and x. The study of this function is much more complex for negative b especially where  x <1.
f: R^{+} . R^{+} > R^{+} More rigorously (or mathematically) we have to found any pair (b, x_{0}) Rigorously, we have to consider only strictly positive value of the real b and x with b not equal to unity. Real vs complex plan considerations In the domain of real definition the two given function f and g are tangent is their graphs plotted as y = f (x) and y = g (x) cross each other only one time at an intersection point x_{t} and that the slope of the two curves is identical at this point so that a tangent line passing at this intersection point can be draw. This graphical situation can be express as a system of two nonlinear equations:
System of equation:
By substituing expression of given in 1) into equation 2), (eq.) ln(b).ln(x_{t}).x_{t} = 1 where b>0, x_{t}>0, b<>1 and x_{t}<>1. This last equation (eq.) give the relation between the slope of the tangent and the position on the curbe (through the abscisse x). More precisely the relation between the logarithm of the slope and the x position. This equation is a product and, since x > 0, the both logarithms ln(b) and ln(x_{t}) must have the same sign. As a consequence,
From (eq) :
Case 1 : b > 1
In the case b > 1, we have ln(b) > 0, thus equation 1. of the initial system is:
In order to avoid x ^{ x } terms we may use the logarithm of this first equation 1. Knowing sign of ln(b), (since ln(b)>0 as b > 1) the system is transformed to :
Replacing any occurrence of ln(b) in 1. :
Posing y = ln(x), we get :
Again, we have to take care of the sign of ln(y) and define three domain :
The only solution in the x > 1 (and b > 1) range was found at x = e where y = ln e is exactly one and ln b = 1/e.
Case 2 : b < 1
In the case 0 < b < 1, we have ln(b) < 0, thus equation 1. of the initial system is :
Since ln x < 0 and ln b < 0, we get :
Replacing any occurrence of ln(b) in 1. :
Posing y = ln(x), we get :
The only solution in the 0 < x < 1 (and 0 < b < 1) range was found at x = 1/e where y = ln(1/e) is exactly one and ln b = e.
Slope of the tangent is 1 (315°). Conclusion: The only two values of b so that f(x) = b^{x} and g(x) = log_{b}(x) are tangent, considering only real space graph, have been found at positions (x,y) = (e,e) and (x,y)=(1/e,1/e) where b = e^{1/e} and b = e^{e} respectively.
Edited: 3 Mar 2009, 9:00 a.m. after one or more responses were posted
Re: OT Weekend Challenge  Gerson W. Barbosa  03032009 Nice work!
Quote: That is, b=1/e^{e} and b=e^{1/e} Looking at your indepth analysis of the problem, I wonder how I was able to find at least one of the solutions :) Gerson.

Re: OT Weekend Challenge  C.Ret  03032009 Thank you, I correct my conclusion from this typo (it is hard not to be lost in the cut/paste composition due to the numerous [italic] and other formatting flags.
Re: OT Weekend Challenge  Chuck  03042009 Wow, excellent analysis C.Ret, and nice graph Gerson. Here was my solution when I came across the problem last week.
Since f(x)=b^x and g(x)=log(b,x) are inverse functions we know that they'll be The source I found the problem in incorrectly assumed that the function had to be tangent to y=x (because they are inverse functions) meaning they have to have a slope of 1, hence only one solution, and not the more interesting 1/e^e base. Fun problem, and thanks to all who played along. CHUCK
EDIT Addendum   Why I find b=1/e^e more interesting is that the functions are not only tangent at x=e^1 (look at Gerson's graph), but also intersect and have the same concavity (not equal to zero). This is amazing! Try to sketch two curves that are tangent, intersect, and have a concavity not equal to zero (or at least one with 0 concavity) . My limited artistic ability this quite challenging. Edited: 4 Mar 2009, 1:59 a.m.
