How do I find $x$ in the equation $x - \ln x = 1.9$?

Next I have $x - 1.9 = \ln x$

we are learning about logarithm but as I tried to take $\ln$ of both side it leads me to nothing.

Could you help me with finding $x$ please?

  • $\begingroup$ What level of math are you in? $\endgroup$ – Shahar Apr 27 '14 at 23:12
  • $\begingroup$ haha Math HL last grade... you think it is too hard for me or it is too easy for me? :D $\endgroup$ – lilludzia Apr 27 '14 at 23:16
  • $\begingroup$ The solution to this is not algebraic and cannot be expressed by elementary operators. You'll need to know the Lambert W function (it may potentially be even more complicated since you have a quotient, not a product) for an exact result, but you can approximate it without that function. $\endgroup$ – Fengyang Wang Apr 27 '14 at 23:21
  • $\begingroup$ @lilludzia Too hard... This isn't very easy to solve. $\endgroup$ – Shahar Apr 27 '14 at 23:36
  • $\begingroup$ You may try solving it using numerical methods : bisection method, newton raphson method, etc. $\endgroup$ – Gaurav May 3 '14 at 19:18

What you can do is graph $$x - \ln x = 1.9 \iff x - 1.9 = \ln x$$

and see (approximate) where the line $y_1 = x-1.9$ intersects $y_2 = \ln x$. The point(s) of intersection is the solution(s), which can only be approximated.

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  • $x \approx 0.1789$,
  • $x\approx 2.9979$.

The solution to this is transcendental (i.e. not algebraic) and so cannot be expressed with elementary operators. Higher-level functions are needed to find an exact answer. Given your mathematics level, these have yet to be introduced.

Maybe the question was to approximate the answer, in which case, notice that 3 is slightly larger than $e=2.71$, so $\ln{3}$ should be slightly larger than 1. But $3-1.9$ is also slightly larger than one, so one of the real solutions is near $x=3$. So $x \approx 3$.

There is one other real solution $0 \lt x \lt 1$ where $\ln{x}$ is sufficiently negative for the left side to approximate 1.9. Taking $x=0$ as an approximation, we have $\ln{x}=-1.9$, so $x={1 \over e^2}$ is a decent approximation. $e^2$ is around 7 so we can say $x \approx {1 \over 7}$.

Any better approximations will probably require a calculator.

Wolfram Alpha tells me that my approximation of 3 is very close (actual root to six significant figures 2.99792), but $1 \over 7$ is quite far off (actual root to six significant figures 0.178863).

This problem can be solved given the Lambert W function. Notice that the equation can be rephrased as $\ln{x}-x=-1.9$ and hence $x e^{-x} = e^{-1.9}$ or $-x e^{-x} = -e^{-1.9}$. The $W(k)$ function gives the solution to equations of the form $x e^x = k$, so the solution is $-x = W(-e^{-1.9})$ hence $x = -W(-e^{-1.9})$. Since $W$ is multivalued, this should give both real solutions.

  • $\begingroup$ The first method I guess you were using the "trial and error" yes? How about this Lambert W Function. How could you reverse it from ln x - x = -1.9 to x e^(-x) = e ^(-1.9) ? This appears to me quite difficult to handle... $\endgroup$ – lilludzia Apr 28 '14 at 0:30
  • $\begingroup$ Rewrite as $e^{\ln{x}-x}=e^{-1.9}$. Then use $e^{a+b}=e^a e^b$. $\endgroup$ – Fengyang Wang Apr 28 '14 at 0:44
  • $\begingroup$ @lilludzia Well, the "trial and error" can be formalized into bisection, which is a pretty standard way of solving equations numerically. $\endgroup$ – orion Apr 28 '14 at 6:33

$$x-\ln x=a\iff e^{x-\ln x}=e^a\iff\frac{e^x}{e^{\ln x}}=e^a\iff\frac{e^x}x=e^a\iff\frac x{e^x}=\frac1{e^a}\iff$$

$$\iff xe^{-x}=e^{-a}\iff-xe^{-x}=-e^{-a}\iff-x=W(-e^{-a})\iff x=-W(-e^{-a}),$$ where W is Lambert's function.


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