How do you solve it?

Geometry Level 3

Find the number of real values of $x$ satisfying $e^{x}=1+\sin x$ .

0 There are infinitely many solutions 1 3 2

2 solutions

Hung Woei Neoh
May 28, 2016

This question is easy if you know the graphs of the equations $y=e^x$ and $y=1+\sin x$ , as shown below:

$Graph of $y=e^x$ and $y=1+\sin x$$ Graph of $y=e^x$ and $y=1+\sin x$

The intersection points of the two graphs will be the solutions to the equation $e^x = 1+ \sin x$ . We can see that for non-negative values of $x$ , or $x \geq 0$ , there is only one solution, which is $x=0$

However, for negative values of $x$ , or $x<0$ , there are infinitely many solutions. This is because for the domain $(-\infty,0)$ , the range of $e^x$ is $(0,1)$ . The graph of $y=1+\sin x$ will cycle continuously in the range $[0,2]$ as shown in the graph above, and intersect the graph of $y=e^x$ repeatedly as it goes up to $2$ and goes down to $0$

Therefore, the number of solutions to this equation is $\boxed{\text{Not Finite}}$

+1 for the graph ... nice solution

Sabhrant Sachan - 5 years ago

I graphed $e^{x}-1-sinx=0$ and got two solutions. I took the x-intercepts.

Hana Wehbi - 5 years ago

You probably graphed it wrongly. Google the equation and have a look at the graph plotted.

Anyway, it's easier to imagine if you graph $y=e^x$ and $y=1+\sin x$ separately instead, and take the intersections of the two graphs

Hung Woei Neoh - 5 years ago

Aditya Narayan Sharma
May 29, 2016

$\displaystyle x=ln|sinx+1|$ For logarithm to exist in real field we must have $-1<sinx\le1$ . Since there are infinitely many x which satisfy these conditions we are done.

Your solution has errors:

It should be $x= \ln |\sin x + 1|$
You only proved that $\ln | \sin x+1|$ is defined. You did not prove that it equals to $x$

Hung Woei Neoh - 5 years ago

Ofcourse it is 'ln' . Anyways for the second part let me give you a solution ,

For very small $x$ , $sinx\approx x$ so we have $e^{x+1}=e(x+1)\implies -(x+1)e^{-(x+1)}=-e^{-1}$

From here recall Lambert W function. $\displaystyle W(-(x+1)e^{-(x+1)})=-(x+1)$

So taking Lambert W on both sides we obtain , $\displaystyle x = -W(-e^{-1})-1$ This would yield solutions.

The graph part straight away shows that the periodic nature is responsible for infinite solutions.

Aditya Narayan Sharma - 5 years ago

I thought $\log$ represents $\log_{10}$ ???

Hung Woei Neoh - 5 years ago

@Hung Woei Neoh – No No you are right about the first part. I've changed it. It was a typo. I explained the second part.

Aditya Narayan Sharma - 5 years ago

How do you solve it?

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