# Contest math question about logarithmic and exponential functions

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 $$1.4 however this doesn't seem to correspond with any value one might expect either for example $$\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)

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 $$g(x) = e\ln(x)$$

$$f(x)=k^x. g(x)=\log_k (x) = \frac{\ln x}{\ln k}$$

They intersect: $$k^x=\ln x/\ln k$$

Their slopes match: $$\ln k \cdot k^x = \frac{1}{x \ln k}=1$$

$$\implies x= 1/\ln k$$

$$\ln k \cdot e^{x \ln k}=1$$

$$\implies \ln k = 1/e\implies k=e^{1/e}\approx 1.444668$$

$$\ln k\cdot e^ {x \ln k}=\frac{1}{x \ln k}$$

$$x \ln k = W(1/\ln k)$$

$$\ln xe^{\ln x}=1/\ln k$$

$$\ln x = W(1/\ln k)$$