# For what number $n$ does the equation $\log_n{x} = n^x$ have exactly one solution?

I'm looking for some number where the graphs of $\log_n{x}$ and $n^x$ intersect at one point - that is, the equation has exactly one solution.

Can someone help me out with this? Based on trial and error I found that it's close to 1.44, but I couldn't find an exact value, even by solving algebraically.

• I'm curious where this came up? I was looking at a similar equation last night. – goblin Jun 20 '18 at 14:46
• @goblin I was screwing around during an exam and it piqued my interest! – Sam Carpenter Jun 21 '18 at 2:11

$x\mapsto \log_n(x)$ and $x\mapsto n^x$ are each other's inverses, so if we're looking for an $n$ where they just touch each other, we're looking for an $n$ where one of them has the line $x=y$ as a tangent.
Set $\alpha = \ln n$; then one of the graphs is $f(x)=e^{\alpha x}$. The tangent condition is that for some $x_0$ we have $$f(x_0) = x_0 \qquad\text{and}\qquad f'(x_0) =1$$ or, concretely $$e^{\alpha x_0} = x_0 \qquad\text{and}\qquad \alpha e^{\alpha x_0} =1$$ Combining these, we find $\alpha x_0 = 1$, so $x_0 = 1/\alpha$, which we can plug back into the second equation to get $$\alpha e^1 = 1$$ which implies $\alpha = e^{-1}$ and thus $n = e^{\alpha} = e^{e^{-1}}$.