Show that the equation $\cos(x) = \ln(x)$ has at least one solution on real number I have question 

Q
  Show that the equation $\cos (x) = \ln (x)$ has at least one solution on real number.

to solve this question by using intermediate value theorem
we let $f(x)=\cos (x)-\ln (x)$
we want to find $a$ and $b$
but what i should try 
to get $f(a)f(b)<0$
I means 
$f(a)>0$
$f(b)<0$
thanks
 A: Hint: $\cos$ is bounded whereas $\ln$ is increasing with $\lim\limits_{x\to 0^+} \ln(x) =- \infty$ and $\lim\limits_{x \to + \infty} \ln(x)=+ \infty$.
A: Plot the graph and you will see immediately.
A: As $-1 \leq \cos x \leq 1$ we have $$-1 -\ln x \leq f(x) \leq 1 - \ln x$$ for all $x > 0$.
To use the intermediate value theorem to show that $f$ has a zero, you are looking for values $a, b > 0$ such that $f(a) > 0$ and $f(b) < 0$. 
As $f(x) \geq -1 - \ln x$, we just need to choose $x$ so that $-1 - \ln x > 0$, and this will be our value $a$. Rearranging (and using the fact that $x \mapsto e^x$ is a strictly increasing function) we need to choose $x$ such that $x < e^{-1}$. So let $a = e^{-2}$. Then $$f(a) \geq -1 -\ln a = -1 - \ln e^{-2} = -1 - (-2) = 1 > 0.$$ That is, $f(a) > 0$.
Using the fact that $f(x) \leq 1 - \ln x$, you should be able to find a value $b$ such that $f(b) < 0$ by a method similar to the above. If you are struggling, I have included the details below, but give it a try yourself first.

 As $f(x) \leq 1 - \ln x$, we just need to choose $x$ so that $1 - \ln x < 0$, and this will be our value $b$. Rearranging (and using the fact that $x \mapsto e^x$ is a strictly increasing function) we need to choose $x$ such that $x > e$. So let $b = e^2$. Then $$f(b) \leq 1 -\ln b = 1 - \ln e^2 = 1 - 2 = -1 < 0.$$ That is, $f(b) < 0$.

A: Think of the graphs of $\ln(x)$ and $\cos(x)$, for $x>0$. The graph of $\cos(x)$ just wiggles back and forth between $-1$ and $1$. On the other hand, the graph of $\ln(x)$ starts out down near $-\infty$, and then increases up to $+\infty$. So, these two graphs must cross each other. Where they cross, we have a solution to the equation.
What I described is just a modified form of the Intermediate Value theorem. If you want to apply the Intermediate value theorem on an interval $[a,b]$, you need to choose $a$ and $b$ so that $f(a)$ and $f(b)$ have opposite signs. Choose $a$ to be a little bit bigger than zero. Say $a = 0.01$. Then $f(a) > 0$. Choose $b$ to be very large. Say  $b = 100$. Then $f(b) < 0$.
As other people said, draw the graphs.
