Can $\ln(x) - \ln(x+1) = 2$ be solved?

Question

I am trying to solve the equation $$ \ln(x) - \ln(x+1) = 2. $$

Using laws of logarithms, the left-hand side can be rewritten as $$\ln\left( \frac{x}{x+1} \right) = 2.$$

Then, by definition of $\ln$:

$$\mathrm{e}^2 = \frac{x}{x+1}.$$

Solve for $x$ from there to get:

$$x = \frac{\mathrm{e}^2}{1 - \mathrm{e}^2}.$$

Which means $x$ is negative for $\ln (x)$ is undefined. Therefore, $\ln (x) - \ln (x+1) = 2$ has no solution.

I have confirmed this by plotting $\ln (x) - \ln (x+1)$, which makes it obvious. Yet I find websites that claim the above is a solution. Not sure I understand why.

Answer 1

Your equation is equivalent to $$\begin{cases} \ln(x)-\ln(x+1)=2\\ x>0 \end{cases}$$ which is also equivalent to $$\begin{cases} x=\dfrac{e^2}{1-e^2}\\ x>0 \end{cases}$$ which is not possible since the second condition is not fulfilled.

Edit: Here the equation is taken over real Numbers (as OP asked) in this case there is no solution. By the way it admit solution over complex numbers.

Answer 2

Assuming that $x>0$,

$$\ln \left(\frac {x}{x+1}\right)=2$$

$$\frac {x}{x+1}=e^2$$

$$x=\frac {e^2}{1-e^2}<0$$

Thus there is no solution.

Answer 3

$ln\frac{x}{x+1}=2$

$\frac{x}{x+1}=e^2$

$x=e^2 \times (x+1)$

$-e^2=e^2 \times x - x$

$-e^2=(e^2 -1) \times x$

$x=\frac{e^2}{1-e^2}$

Hence, $x$ is negative, and it's natural log doesn't exist.

Answer 4

The existence (or non-existence) of solutions to the equation $$ \ln(x) - \ln(x-1) = 2 $$ depends on how we choose to define the logarithm function. Specifically, it depends on how the domain is chosen.

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In high school mathematics, the logarithm function is generally defined as the inverse of the exponential function. That is, $$ \ln(x) = y \iff x = \mathrm{e}^y,$$ where the exponential function $y \mapsto \mathrm{e}^y$ is defined for all real numbers, and is always positive. This implies that $$ \ln : \mathbb{R}_{\ge 0} \to \mathbb{R}, $$ that is, the domain of the logarithm function is the positive real numbers. Under this definition of the logarithm function, if we assume that $$ \ln(x) - \ln(x-1) = 2 $$ has a real solution, then we must have $x > 0$ (as $\ln(x)$ will be undefined otherwise), and we must have that $x > -1$ (as $\ln(x+1)$ will be undefined otherwise; note that this condition is redundant in light of the previous condition). Thus \begin{align} \ln(x)-\ln(x+1) = 2 \land x > 0 &\iff \ln\left( \frac{x}{x+1} \right) = 2 \land x > 0 \\ &\iff \frac{x}{x+1} = \mathrm{e}^2 \land x > 0 \\ &\iff x = \frac{\mathrm{e}^2}{1-\mathrm{e}^2} < 0 \land x > 0. \end{align} The last line says that $x$ is both negative and positive, which is impossible, and contradicts the original assumption, which was that the equation $\ln(x) - \ln(x+1) = 2$ had a real solution. Therefore this equation does not have a solution in the real numbers.

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From a slightly more advanced point of view (though still a "mortal" point of view—the relevant theory is taught to undergraduates), the logarithm function can be extended to (most of) the complex plane. There are some technical hurdles which need to be overcome: the exponential function is not one-to-one on the complex plane, and so it does not have an inverse. To obtain an inverse, the exponential function must be restricted to some subset of the complex plane on which it is one-to-one (and therefore invertible). Once the exponential has been restricted to such a domain, the inverse function is a "branch" of the logarithm function. That is, "the" complex logarithm is the inverse of the exponential function after the exponential function has been properly restricted to a domain on which it is invertible (if you have seen trigonometric functions and their inverses, it is a similar story).

The explanation above is quite hand-wavy, in deference to the apparent level of the asker. For the more advanced reader, please note that it is not meant to be a rigorous discussion. For a more rigorous explanation, Wikipedia does okay.

In any event, it is possible to define a branch of the logarithm function such that the original equation has a solution. In that context, every step of the following computation may be justified: \begin{align} \ln(x)-\ln(x+1) = 2 &\iff \ln\left( \frac{x}{x+1} \right) = 2 \\ &\iff \frac{x}{x+1} = \mathrm{e}^2 \\ &\iff x = \frac{\mathrm{e}^2}{1-\mathrm{e}^2}, \end{align} and we get the desired solution.

When a website like Wolfram|Alpha finds a solution to this equation, it is doing so in the context of the complex logarithm. It may be worth noting that this is pointed out by the software: