Here's the question I'm trying to solve:
For what value of $k$ is the function $f(x)=k^x$ tangent to $g(x)=\log_k(x)$?
My Attempt:
Since $f(x)$ and $g(x)$ are inverses of each other and are therefore symmetrical on $y=x$, the tangent point must occur when $y=x$. By setting $f(x)=x$ and $g(x)=x$, we get $x=k^x$ and $x=\log_{k}x$ but these two equations are pretty much the same since $x=k^x \Rightarrow x=\log_{k}x$ so I can't find any unique solution. Most other approaches I've tried pretty much get me to the same two equations.
I've tried taking the derivative of both functions which have to be 1; $f'(x)=\ln(k)k^x=1$ and $g'(x)=\frac{1}{\ln(k)x}=1$ and since previously $k^x=x$ we have $x \ln(k)=1$ but we get the same issue where we get two equivalent equations.
After graphing it and trying to reverse engineer the answer I found that $1.4<k<1.5$ however this doesn't seem to correspond with any value one might expect either for example $\frac{e}{2}\approx 1.36$ or $\sqrt{e}\approx1.67$ or $\sqrt[3]{e}\approx 1.39$.
Note that this is a math competition question so should in theory (for this competition) be doable in about 5 minutes so I'm sure there's a trick to it that I have't figured out. (maybe Lambert W although I doubt it)
Best Answer
Here it goes:
As $f(x)$ and $g(x)$ are inverse of each other, for them to be tangent at one point, The slope at their point of intersection must be "$1$"
Let them meet at point $(x_1,y_1)$
hence, $f(x_1)=g(x_1)=x_1$ because, they both are also tangent to line $y=x$
putting the values
$$ k^{x_1} = \frac {\ln(x_1)}{\ln(k)}=x_1 \tag{1} $$
now derivative of both curves at $x=x_1$ must equal "$1$"
$$ f'(x_1) = g'(x_1) = 1 $$
substituting values
$$ k^{x_1}\ln(k) = \frac{1}{x_1\ln(k)} = 1 \tag{2} $$
using this, $$ x_1 = \frac{1}{\ln(k)} \tag{3} $$
substitute
$x_1 = \frac{1}{\ln(k)}$ in Equation (1)
hence
$$ \frac{\ln\Bigl(\frac{1}{\ln(k)}\Bigr)}{\ln(k)} = \frac{1}{\ln(k)} $$
by cross multiplying terms
$$ \ln\biggl(\frac{1}{\ln(k)}\biggr) = 1 $$
hence
$$ \frac{1}{\ln(k)} = e $$
$$ \ln(k) = \frac{1}{e} $$
which gives
$$ k = e^{\frac{1}{e}} $$
substituting value of "$k$" in function yields
$$ f(x) = e^{\frac{x}{e}} $$ and $$ g(x) = e\ln(x) $$