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 non-obvious 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^{+}
( b , x ) b^{x}
g: |R^{*+}-{1} . |R^{*+} -------> |R
( b , x ) ln x / ln b
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 non-linear equations:
- y = f (x_{t}) and y = g (x_{t})
- s = d/dx f (x_{t}) and s = d/dx g (x_{t})
System of equation:
- f (x_{t}) = g (x_{t})
- f’ (x_{t}) = g’ (x_{t})
- b^{xt} = ln(x_{t})/ln(b)
- ln(b).b^{xt} = 1/( x_{t}.ln(b)
- b^{xt} = ln(x_{t})/ln(b)
- ln(b)^{2}.b^{xt}. x_{t} = 1
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,
- 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 x_{t} = 1 or b = 1.
From (eq) :
ln(b) = 1 / ( ln(x_{t}).x_{t} )
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 x_{t} > 1.
In the case b > 1, we have ln(b) > 0, thus equation 1. of the initial system is:
- b^{x } = ln(x)/ln(b)
- 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 :
- ln(b^{x}) = ln( ln(x)/ln(b) )
- ln(b) = 1 / ( ln(x).x )
- x.ln(b) = ln(ln(x)) - ln(ln(b))
- ln(b) = 1 / ( ln(x).x )
Replacing any occurrence of ln(b) in 1. :
- x / ( ln(x).x ) = ln(ln(x)) + ln(ln(x).x)
- ln(b) = 1 / ( ln(x).x )
- x / ln(x) / x = ln(ln(x)) + ln(ln(x)) + ln(x)
- ln(b) = 1 / ( ln(x).x )
- 1 / ln(x) = 2.ln(ln(x)) + ln(x) where x > 1
- ln(b) = 1 / ( ln(x).x )
Posing y = ln(x), we get :
- 1 / y = 2.ln(y) + y where y > 0
- 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 x_{t} = exp(1) = e.
- 1 / 1 = 0 + 1
- ln(b) = 1 / ( 1.e^{1} )
Equation 1. is OK from 1 = ln(x_{t}) we get intersection coordinate and b parameter :
- x_{t} = e which is > 1
- ln b = 1 / e
- x_{t} = 2.7183
- 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 < x_{t} < 1.
In the case 0 < b < 1, we have ln(b) < 0, thus equation 1. of the initial system is :
- b^{x } = ln(x)/ln(b)
- ln(b) = 1 / ( ln(x).x )
- ln(b^{x}) = ln( ln(x)/ln(b) )
- ln(b) = 1 / ( ln(x).x )
Since ln x < 0 and ln b < 0, we get :
- x.ln(b) = ln(-ln(x)) - ln(-ln(b))
- ln(b) = 1 / ( ln(x).x )
Replacing any occurrence of ln(b) in 1. :
- x / ( ln(x).x ) = ln(-ln(x)) + ln(-ln(x).x)
- ln(b) = 1 / ( ln(x).x )
- x / ln(x) / x = ln(-ln(x)) + ln(-ln(x)) + ln(x)
- ln(b) = 1 / ( ln(x).x )
- 1 / ln(x) = 2.ln(-ln(x)) + ln(x) where 0 < x < 1
- ln(b) = 1 / ( ln(x).x )
Posing y = -ln(x), we get :
- 1 / y = y - 2.ln(y) where y > 0
- ln(b) = 1 / ( ln(x).x )
- if y = 1 then ln y = 0 an acceptable solution exist at position 1/e where 1 = -ln(x_{t})[\italic]
- [italic] x_{t} = 1/e = .3679
- 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) = 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