Interception of exp. and log. curves

[f1player]

Consider the curves y = e^x and y = ln(ax)

For what value of a does the first curve touch the second?

Here's what I did:

e^x = ln(ax)
e^x = lna + lnx
e^x - lnx = lna

a = e^(e^(x) -lnx) Equation 1

Also: Derivatives of both curves equal each other, as tangents will have same gradient

Therefore: y' = e^x and y' = 1/x

That is: e^x = 1/x Equation 2

So now I have 2 equations and 2 unknowns (a and x) but I still cant find a way to solve for a.

Anyone able to help???

By the way, the value for a is about 10.28 or 10.29, which can be seen when graphed on a calculator, but how would you solve algebraically?

 

[soroban]

Hello, f1player!

Consider the curves \(\displaystyle y\,=\,e^x\) and \(\displaystyle y\,=\,\ln(ax)\)

For what value of \(\displaystyle a\) does the first curve touch the second?

Your reasoning and your work are absolutely correct.
However, we cannot solve those equations algebraically.
The best we can hope for is an approximation (which you got by graphing).

Here's my view of the problem . . .

\(\displaystyle y\,=\,e^x\) and \(\displaystyle y\,=\,\ln x\) are inverse functions.
\(\displaystyle \;\;\)They are symmetric to the 45^o-line and have no points in common.

                          |      *
                          |             /
                          |     *     /
                          |    *    /
                          |  *    /
                          *     /       *
                      *   |   /    *
              *           | /   *
        - - - - - - - - - / - * - - - - - - - - 
                        / |
                      /   | *
                    /     |
                  /       |*

However, \(\displaystyle y\:=\:\ln(ax)\:=\;\ln(x)\,+\,\ln(a)\)

Hence, \(\displaystyle y\,=\,\ln x\) is translated upward \(\displaystyle \ln a\) units.

Therefore, for the right choice of \(\displaystyle a\), we can make the curves "kiss".


But this got me no closer to the answer . . .