Solving equation involving logarithm

Question

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?

Answer 1

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.

$$ $$

Solutions:

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

Answer 2

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).

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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.

Answer 3

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

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