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My Mathematics Teacher gave me the following problem :

Find all the possible values of $x$ if $\ln(x) = \sin(x)$.

I tried graphing both $\ln(x)$ and $\sin(x)$. I found that they intersect at a single point $\approx (2.22, 0.8)$. However, I am unable to find a "mathematical" solution to this problem using the elementary properties of trigonometry and logarithms. I tried some Taylor expansions, but that did not help. Can anybody provide a hint as to what should be done ?

Also, what can be the possible values of $x$ is the problem is generalized as $\sin(x)=\log_b(x) ? $ One simple observation is that $0

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    You can only find an approximate value. – 2017-02-02
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    @Mercy King, OP wants a mathematical explanation of why the curves intersect only at a single point. – 2017-02-02
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    Since you know that there is a unique solution, you can find an initial bracket, say $[1, e]$ that contains the solution. Let $f(x) = \ln x - \sin x$, then $f(1) <0, f(e) >0$. Choose the midpoint of the interval and evaluate $f$ at that point. Split the interval into two parts. If $>0$ choose the left interval and similarly if $<0$ choose the right interval. Repeat until the error is small enough (the maximum error halves at each stage). – 2017-02-02

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As said in answers and comments, the solution requires numerical methods.

Since you graphed the functions, you noticed that they intersect close to $x=2$. Then, in order to approximate the solution, you could have used Taylor expansions (as you though about) but building them around $x=2$.

You have $$\log(x)=\log (2)+\frac{x-2}{2}-\frac{1}{8} (x-2)^2+\frac{1}{24} (x-2)^3+O\left((x-2)^4\right)$$ $$\sin(x)=\sin (2)+(x-2) \cos (2)-\frac{1}{2} (x-2)^2 \sin (2)-\frac{1}{6} (x-2)^3 \cos (2)+O\left((x-2)^4\right)$$ Limited to first order, we then have $$\log(x)-\sin(x)=(\log (2)-\sin (2))+(x-2) \left(\frac{1}{2}-\cos (2)\right)+O\left((x-2)^2\right)$$ leading to the solution $$x_{(1)}=\frac{2 (-1+\log (2)-\sin (2)+2 \cos (2))}{2 \cos (2)-1}\approx 2.23593$$ Limited to second order,we have $$\log(x)-\sin(x)=((\log (2)-\sin (2))+(x-2) \left(\frac{1}{2}-\cos (2)\right)+(x-2)^2 \left(\frac{\sin (2)}{2}-\frac{1}{8}\right)+O\left((x-2)^3\right)$$ and solving the quadratic would give $$x_{(2)}\approx 2.21872$$ while the "exact" solution given by Newton method would be $$x\approx 2.21911$$

Edit

May be, it would be nicer to build Taylor expansion around $x=\frac{2\pi}3$. This would give $$\log(x)-\sin(x)=\left(\log \left(\frac{2 \pi }{3}\right)-\frac{\sqrt{3}}{2}\right)+\frac{(3+\pi ) }{2 \pi } \left(x-\frac{2 \pi }{3}\right)+\left(\frac{\sqrt{3}}{4}-\frac{9}{8 \pi ^2}\right) \left(x-\frac{2 \pi }{3}\right)^2+O\left(\left(x-\frac{2 \pi }{3}\right)^3\right)$$ Limiting to first order, the solution would be $$x_{(1)}=\frac{\pi \left(6+3 \sqrt{3}+2 \pi -6 \log \left(\frac{2 \pi }{3}\right)\right)}{3 (3+\pi )}\approx 2.22408$$ Using the second order, this would give $$x_{(2)}\approx 2.21901$$

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I can't think of anything other than a numerical method to find the answer.

However, you can use purely mathematical methods to prove that there is only one answer.

$\ln x < \sin x$ for all $x\le \frac {\pi}2$

$\ln x > \sin x$ for all $x\ge \frac {3\pi}2$

IVT says that there is at least one point in $(\frac {\pi}2, \frac {3\pi}2)$ where $\sin x = \ln x$

Since $\frac {d}{dx} \sin x < 0$ and $\frac {d}{dx} \ln x > 0$ for all $x$ in this inteval there is only one point in the inteval where $\sin x = \ln x$

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    What does "IVT" mean here ??? – 2017-02-02
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    Intermediate Value Theorem – 2017-02-02
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    OK Thanks. What about the generalized question ???? – 2017-02-02
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    For large values of $b.$ there could be multiple solutions. e.g. $\log_{10} x = \sin x$ has $3$ solutions. – 2017-02-02
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Observe that $\ln$ is an unbounded, strictly increasing function, and that $\sin$ is periodic with range $-1 \le \sin(x) \le 1$. So the only area where there can be such a solution is in the area $1/e\le x \le e$, the area where $-1 \le \ln(x) \le 1$.

We can then break this area up into three parts:

  • $1/e \le x < 1$, where both $\ln(x)$ and $\sin(x)$ are increasing and $\ln(x)<0$ but $\sin(x)>0$, so there cannot possibly be an intersection there,
  • $1 < x < \pi/2$, where both are increasing and positive; however, in this area, $\ln(x)$ ranges from $0$ to about $0.45$, and $\sin(x)$ ranges from about $0.82$ to $1$, so there still cannot possibly be an intersection there,
  • and finally $\pi/2 < x \le e$, where $\ln(x)$ is increasing and $\sin(x)$ is decreasing; this is where they cross, and since the slopes are now opposite signs, there must be only one.

Once we've done this, we can use standard numerical root-finding techniques to locate the root. There's no hope in finding an analytic solution.