Show that for a ≥ e, there exists at least one line that is simultaneously tangent to both f(x) = ax and
g(x) = loga x.
Bottom line: Yes - we prove below that there are solutions
for a>e, as an example, for a=3, and we plot one.
I advanced the constraints to 3 eqn.s and 3 unknowns.
from previous post:
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f(x) = a^x,
f'(x)=a^x ln a
g(x) = log_a x = ln x / ln a
g'(x) = 1 / (x ln a)
The line has slope m and is tangent to
f and g at x1 and x2 respectively with
slope m, so:
f'(x1) = g'(x2)
a^x1 ln a = 1 / (x2 ln a) [eq. 1]
[g(x2) - f(x1)] / ( x2 - x1) = f'(x1)
(ln x2 / ln a - a^x1) / (x2 -x1) = a^x1 ln a [eq. 2]
[g(x2) - f(x1)] / ( x2 - x1) = g'(x2)
(ln x2 / ln a - a^x1) / (x2 -x1) = 1/ (x2 ln a) [eq. 3]
But, there are really only two independent equations here.
newly added:
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The y-intercept "b" of the line can be computed from each
function to add one more equation:
Using points (x1,y1) and (x2,y2) on f and g, respectively:
b = b
y1 - m x1 = y2 - m x2
f(x1) - x1 f'(x1) = g(x2) - x2 g'(x2)
a^x1 - x1 a^x1 ln a = ln x2 / ln a - x2 [1/ (x2 ln a)]
a^x1 - x1 a^x1 ln a = ln x2 / ln a - 1/ ln a [eqn. 4]
-----------------------------------------------------------
Putting equations 1, 2, and 4 together, and putting in a
constant a=3, Wolfram Alpha gives us a relationship*
between x1 and x2 that satisfies the constraints.
3^x1 (1 - x1 ln a) = [(ln x2) - 1] / ln 3
A random pair in this relationship is (x1,x2) = (1.400, 0.171)
This gives the line
y = 3^1.40 (ln 3) x + [3^1.40 - 1.40 3^1.40 ln 3]
which we plot here:
https://www.desmos.com/calculator/4vcsqqo00e
* Wolfram Alpha actually give three distinct possible
relationships between x1 and x2 -
three families of solutions for a=3
where we now have plotted x1 against x2 on the
x,y plane.
Edited on January 20, 2025, 10:50 am
a bit more...
