# How to solve $3^{x-1} \ln(x)=x^{x}-1$ without simply guessing $x=1$?

This is an algebra question, in spite of looking like a Differential Calculus one.

If $$\theta$$ is the angle of intersection of the curve $$y=3^{x-1}\ln(x)$$ and $$y=x^{x}-1,$$ then $$2\cos(\theta)$$ is equal to?

This question was given by my teacher and I presume it’s a common question found in most books. Naturally the approach to such a question would begin by finding the common point of intersection and evaluating the slopes. But here lies my problem,

How do you solve this equation? $$3^{x-1} \ln(x)=x^{x}-1$$

When I consulted my teacher, he said that the only approach (my teacher being one who teaches for competitive examinations which are objective based) would be to observe that the equation has an exponential, logarithmic and a [function to a function] and presumably can't be solved. Hence you'll have to observe that 1 is the only solution to this equation.

But, since this is an objective examination [and I'm not good as guessing math], you have to decisively prove that one is the only solution and your question could be wrong if there are more than one solutions (since that would indicate that there are more than one point of intersections giving multiple slopes and angles). Edit: I just found this answer to an identical question regarding that, What is the cosine of angle of intersection of following functions?

But I have 2 issues with it,

1. It doesn't address the point of intersection problem I'm facing
2. I don't get the tangent vector idea from this.
• $x=1$ is one solution of your equation. – Dr. Sonnhard Graubner Oct 13 '19 at 12:48
• @Dr.SonnhardGraubner if you mean to say that "1 is the solution", how did you get that? If you wish to convey that " 1 is the one solution", well I've stated yet and also stated that if there are more than one solutions, we'll be dealing with many points of intersection. How do we tackle that issue? – Shreyas JV Oct 13 '19 at 13:07
• Curiously, according to Wolfram Alpha, $x^x - 1 \geqslant 4.25^{x-1}\ln(x)$ for all $x > 0$, but (i) the difference (LHS - RHS) has a local maximum at $x \bumpeq 1.34661$, and (ii) the inequality doesn't hold for all $x > 0$ if $4.25$ is replaced by $4.26$. – Calum Gilhooley Oct 14 '19 at 16:36
• Wolfram Alpha can't calculate the minimum of $\sqrt[x-1]{\frac{x^x-1}{\ln(x)}}$ for $x > 0$ (although it calculates the exact limit $e^{3/2} \bumpeq 4.48169$ at $x = 1$), but the graph appears to descend to a global minimum of just over $4.259$ at $x\bumpeq1.584$, then rise to $+\infty$. – Calum Gilhooley Oct 14 '19 at 17:20

Note that for $$x\ge3$$, from this post, $$3^{x-1}\ln x\le3^{x-1}\cdot\frac{x-1}{\sqrt x}<3^{x-1}\sqrt x since $$3^{x-1} The interval $$(0,3)$$ can be dealt with by using $$(x-1)/\sqrt x$$ directly, but the algebra is less straightforward -- will add details shortly.