You are asking for the other positive root of the equation $f(x)=0$, where $f(x)=2x+\ln(1-0.1x ).$ The solution cannot be expressed exactly by using simple combinations of familiar functions and constants. But if we think of the problem as a practical problem, we can obtain an answer that is as accurate as we need.
There are many possible approaches. For example,if you have access to a graphing calculator, you can ask it to graph the curve $y=f(x)$. Then you can read the second root off the display. To get additional accuracy, use the "zoom" feature.
Various graphing programs are available, for a wide variety of computer platforms. This includes a number of free programs.
Wolfram Alpha can be freely used. It will graph for you, solve equations for you, and much much more.
Finally, we can attack the problem ourselves. It is useful to have access to a simple scientific calculator, or to a calculator program. (Such a program is probably already on the computer that you use.)
Many root-finding algorithms have been developed. A useful one to know is the Newton-Raphson Method. An often even better one is the Secant Method. Once these methods get a good smell of the root, they usually home in on it with remarkable efficiency.
We will use a less fancy but still systematic method, often called the Bisection Method. It will be best if you do the calculations below with your own calculator.
We need two starting points, one of them smaller than the root, and the other bigger than the root. Preferably these starting points are close to each other.
Our curve $y=2x+10\ln(1-0.1x)$ goes up and then down. Have we hit the ground by the time $x=7$? Let's calculate $f(7)$. No, $f(7)$ is about $1.96$. Now let's calculate $f(9)$. We get $f(9)\approx -5.0$.
So far we have $f(7)>0$, and $f(9)<0$. Since our function $f$ is continuous, there is a root between $x=7$ and $x=9$.
Now look at the point halfway between $7$ and $9$, namely $8$. So we bisect the interval. Calculate $f(8)$. We get $-0.09$. So $x=8$ is awfully close to a root. Not a very impressive kick!
Now I will be a little silly, and continue like a machine. We have $f(7)>0$, and $f(8)<0$, so look at the point halfway between, namely $7.5$. We have $f(7.5)>0$. Since $f(8)<0$, our root is between $7.5$ and $8$. Look at the point halfway in between, namely $7.75$.
We have $f(7.75)>0$, so our root is between $7.75$ and $8$. Look at the point halfway in between, namely $7.875$. We find that $f(7.875)>0$. So there is a root between $7.875$ and $8$. Go halfway in between, to $7.9375$. We have $f(7.9375)>0$. Go halfway between $7.9375$ and $8$, to $7.96875$. We find that $f(7.96875)<0$. So our root is between $7.9375$ and $7.96875$.
If we need greater accuracy, continue. Each time we go through one cycle of the Bisection Method, our region of uncertainty shrinks by a factor of $2$. After a while, we are close enough for all practical purposes.
Remark: In our case, since $x=8$ is so close to a root, you may have been just intended to just play around for a bit, and notice that $f(8)$ is nearly $0$.
You might want to look up the Newton Method, or the Secant Method, and see how they perform.