Logarithm curves intersection number

Question

$y=\log (x)$ and $y=\frac{1}{x}$ are drawn in a plane. Try and find number of times they intersect for values of $x$ greater than 1.

I equated the two values of $y$. $$\log (x) = \frac{1}{x}$$

$$\log(x)^x=1$$ $$x^x=e$$ Then what?

Answer 1

$f(x) = \log x$ is an increasing function.

$g(x) = \frac1x$ is a decreasing function on the positive domain.

Hence they intersect at most once. Can you show that they intersect? For example, by using intermediate value theorem?

Remark about your attempt:

From $\log x = \frac1x$, we have $x^x = \color{red}{e}$

Also from $\log x^x =1 $ (which is wrong as pointed above), we can't say $x^x= x$,

Answer 2

For $x \ge 1$ let $f(x)= \log(x)-1/x$. Then $f(1)<0$ and $f(e)>0$, hence there is $x_0>1$ such that $f(x_0)=0$. Since $f'(x)>0$ for $x>1$, the equation $f(x)=0$ has exactly one solution.

Answer 3

Consider the function $$F(x)=\log(x)-\frac 1x \implies F'(x)=\frac 1x+\frac 1{x^2} >0 \qquad \forall x >0$$ So, $F(x)$ is an increasing function starting from $-\infty$ so only one root for $F(x)=0$.

By inspection, $F(1)=-1$ and $F(e)=1-\frac 1e >0$ give simple bounds and you can start any procedure for finding the solution.