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OT Weekend Challenge
Message #1 Posted by Chuck on 27 Feb 2009, 2:43 p.m.

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
Message #2 Posted by Allen on 27 Feb 2009, 7:42 p.m.,
in response to message #1 by Chuck

Quote:
Given f(x)=b^x and g(x)=log(b,x) find all possible values for b so that f and g are tangent.

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 mis-read 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
Message #3 Posted by Chuck on 27 Feb 2009, 8:32 p.m.,
in response to message #2 by Allen

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
Message #4 Posted by Namir on 27 Feb 2009, 11:10 p.m.,
in response to message #1 by Chuck

This is not some homework you need help with?

:-)

Namir

            
Re: OT Weekend Challenge
Message #5 Posted by C.Ret on 28 Feb 2009, 2:53 a.m.,
in response to message #4 by Namir

Quote:
This is not some homework you need help with?

It's really like it is the case.

May I suggest to look at e and 1/e.

                  
Re: OT Weekend Challenge
Message #6 Posted by Chuck on 28 Feb 2009, 11:12 a.m.,
in response to message #5 by C.Ret

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
Message #7 Posted by Chuck on 28 Feb 2009, 11:08 a.m.,
in response to message #4 by Namir

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
Message #8 Posted by Gerson W. Barbosa on 28 Feb 2009, 9:54 p.m.,
in response to message #1 by Chuck

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
Message #9 Posted by Namir on 28 Feb 2009, 10:04 p.m.,
in response to message #8 by Gerson W. Barbosa

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
Message #10 Posted by C.Ret on 2 Mar 2009, 8:37 p.m.,
in response to message #9 by Namir

Shall I propose to start over from start:

Given f(x) = bx and g(x) = logb(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 non-obvious graphical approach have to be explained.

The (real) natural logarithms ln x is only define for strictly positive real argument x.

Since logb 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 bx 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+
   (    b    ,  x  )          bx

g: |R*+-{1} . |R*+ -------> |R ( b , x ) ln x / ln b

More rigorously (or mathematically) we have to found any pair (b, x0)

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 xt 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 non-linear equations:

  1. y = f (xt) and y = g (xt)
  2. s = d/dx f (xt) and s = d/dx g (xt)

System of equation:

  1. f (xt) = g (xt)
  2. f (xt) = g (xt)

  1. bxt = ln(xt)/ln(b)
  2. ln(b).bxt = 1/( xt.ln(b)

  1. bxt = ln(xt)/ln(b)
  2. ln(b)2.bxt. xt = 1

By substituing expression of given in 1) into equation 2),

(eq.) ln(b).ln(xt).xt = 1 where b>0, xt>0, b<>1 and xt<>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(xt) must have the same sign.

As a consequence,

  • if a tangent exist with b less than 1, it may only exist at x positions in interval ] 0 ; 1 [
  • if a tangent exist with b more than 1, it may only exist at x positions greater than 1 ( ] 1 ; oo [.
  • no tangent point is possible at xt = 1 or b = 1.

From (eq) : ln(b) = 1 / ( ln(xt).xt )

Case 1 : b > 1 Supposing a tangent of correct slope exist with b > 1, we have to verify that an unique intersection also exist at a point xt > 1.

In the case b > 1, we have ln(b) > 0, thus equation 1. of the initial system is:

  1. bx = ln(x)/ln(b)
  2. ln(b) = 1 / ( ln(x).x )

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 :

  1. ln(bx) = ln( ln(x)/ln(b) )
  2. ln(b) = 1 / ( ln(x).x )

  1. x.ln(b) = ln(ln(x)) - ln(ln(b))
  2. ln(b) = 1 / ( ln(x).x )

Replacing any occurrence of ln(b) in 1. :

  1. x / ( ln(x).x ) = ln(ln(x)) + ln(ln(x).x)
  2. ln(b) = 1 / ( ln(x).x )

  1. x / ln(x) / x = ln(ln(x)) + ln(ln(x)) + ln(x)
  2. ln(b) = 1 / ( ln(x).x )

  1. 1 / ln(x) = 2.ln(ln(x)) + ln(x) where x > 1
  2. ln(b) = 1 / ( ln(x).x )

Posing y = ln(x), we get :

  1. 1 / y = 2.ln(y) + y where y > 0
  2. ln(b) = 1 / ( ln(x).x )

Again, we have to take care of the sign of ln(y) and define three domain :

  • if ln(y) = 0 then the system accept a solution : y = 1 ---> ln(y) = 0 and the system admit an unique intersection xt = exp(1) = e.
    1. 1 / 1 = 0 + 1
    2. ln(b) = 1 / ( 1.e1 )

    Equation 1. is OK from 1 = ln(xt) we get intersection coordinate and b parameter :

    1. xt = e which is > 1
    2. ln b = 1 / e

    1. xt = 2.7183
    2. b =1.4447

    Intersection coordinate are (e,e) and slope of the tangent is 1 (i.e. 45)

  • if ln(y) > 0 then the system accept no solution :

    Since ln is a growing function, ln(y) > 0 imply that y > 1. In this case, the left side of equation (1.) 1/y is less than one and can never be equal to the right side 2.ln(y) + y which has two raisons to be greater than one.

    No solution can be found for any x greater than e (where y >1, x > 1, b >1 ).

  • if ln(y) < 0 then the system accept no solution :

    Since ln is a growing function, ln(y) < 0 intimate that y < 1. In this case, the left side of equation 1. 1/y is more than one and can never be equal to the right side 2.ln(y) + y which has two raisons to be less than one.

    No solution can be found for any x in the interval ] 1 ; e [ (where y <1, x>1, b>1).

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 Supposing a tangent of correct slope exist with 0 < b < 1, we have to verify that an unique intersection also exist at a point 0 < xt < 1.

In the case 0 < b < 1, we have ln(b) < 0, thus equation 1. of the initial system is :

  1. bx = ln(x)/ln(b)
  2. ln(b) = 1 / ( ln(x).x )

  1. ln(bx) = ln( ln(x)/ln(b) )
  2. ln(b) = 1 / ( ln(x).x )

Since ln x < 0 and ln b < 0, we get :

  1. x.ln(b) = ln(-ln(x)) - ln(-ln(b))
  2. ln(b) = 1 / ( ln(x).x )

Replacing any occurrence of ln(b) in 1. :

  1. x / ( ln(x).x ) = ln(-ln(x)) + ln(-ln(x).x)
  2. ln(b) = 1 / ( ln(x).x )

  1. x / ln(x) / x = ln(-ln(x)) + ln(-ln(x)) + ln(x)
  2. ln(b) = 1 / ( ln(x).x )

  1. 1 / ln(x) = 2.ln(-ln(x)) + ln(x) where 0 < x < 1
  2. ln(b) = 1 / ( ln(x).x )

Posing y = -ln(x), we get :

  1. 1 / y = y - 2.ln(y) where y > 0
  2. ln(b) = 1 / ( ln(x).x )

  • if y = 1 then ln y = 0 an acceptable solution exist at position 1/e where 1 = -ln(xt)[\italic]

    1. [italic] xt = 1/e = .3679
    2. b = 0.0660

  • if 0 < y < 1 or y > 1, no solution can be found.

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). Note that the two curve are tangent but still crossing.

Conclusion:

The only two values of b so that f(x) = bx and g(x) = logb(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 = e1/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
Message #11 Posted by Gerson W. Barbosa on 3 Mar 2009, 7:14 a.m.,
in response to message #10 by C.Ret

Nice work!

Quote:
Conclusion:

The only two values of b so that f(x) = bx and g(x) = logb(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 ln b = 1/e and ln b = -e respectively.


That is, b=1/ee and b=e1/e

Looking at your in-depth analysis of the problem, I wonder how I was able to find at least one of the solutions :-)

Gerson.

------------------

Graphmatica plots of y=ln(x)/ln(1/e^e), y=(1/e^e)^x, y=(e^(1/e))^x and y=ln(x)/ln(e^(1/e)) :

                              
Re: OT Weekend Challenge
Message #12 Posted by C.Ret on 3 Mar 2009, 8:56 a.m.,
in response to message #11 by Gerson W. Barbosa

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
Message #13 Posted by Chuck on 4 Mar 2009, 1:09 a.m.,
in response to message #12 by C.Ret

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
tangent along with y=x for some number x=a.  This means, 

a=b^a and a=ln(a)/ln(b) (using a change of base for g(x))

Also, since they are tangent , their derivatives will also be equal at some x=a, or

b^a ln(b) = 1/(a ln(b))

These three equations allow us to solve for a. Using the first two equations (into the third) we get

a ln(b) = 1/ln(a) ln(a) = 1/ln(a) (ln(a))^2 = 1 ln(a) = +/- 1

so, a = e^(+/-1)

Therefore, since b=a^(1/a) we have b=e^(1/e) and b=1/(e^e), giving the two solutions and the nice graphs Gerson made.

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.


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