# 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

Hint: $\cos$ is bounded whereas $\ln$ is increasing with $\lim\limits_{x\to 0^+} \ln(x) =- \infty$ and $\lim\limits_{x \to + \infty} \ln(x)=+ \infty$.

• but what will be the interval [a,b] this i need please thanks Apr 26 '13 at 9:26
• Just take $a$ and $b$ such that $\ln(a)<-1$ and $\ln(b)>1$. Apr 26 '13 at 9:31

Plot the graph and you will see immediately.

• but we must have interxal [a,b] and why you use limit we try some vales that f(a)<0 other one f(b)>0 thanks Apr 26 '13 at 6:37
• @leenaadam, not sure what you mean. I didn't use limit. If you insist using intermediate value theorem, simply let $x=0$ and $x=\frac{\pi}{2}$ will give you want you want.
– Easy
Apr 26 '13 at 6:41
• but if i use 0 we get ln(0) is infinity and we get 1-infinitu can you explain more this part please thanks Apr 26 '13 at 6:49
• @leenaadam, if you think 0 is not defined, you can find a point close to 0 but not exactly is, say 0.1? Actually if you draw the graph, you will understand what happens.
– Easy
Apr 26 '13 at 6:52

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, $f(b) < 0$.

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.