# Logarithm problem that I'm stuck on

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1. Jul 6, 2015

### Rectifier

The problem statement
ln(x) = 5 -x
Solve for x.

The attempt at a solution
$$ln x = 5 - x \\ e^{ln x} = e^{5 - x} \\ e^{ln x} = \frac{e^5}{e^x} \\ x e^x = e^5$$
Here is the place where I get stuck.

2. Jul 6, 2015

### DrDu

There is no analytical solution. You have to solve it numerically, e.g. $x_{n+1}=5-\ln(x_n)$ starting with $x_0=5$.

3. Jul 6, 2015

### Orodruin

Staff Emeritus
In addition to what DrDu said already, the formal solution to the problem is given in terms of the Lambert W function. As indicated in the previous post, you will have to find the solutions to this numerically.

4. Jul 6, 2015

### EM_Guy

$5 = x + ln(x)$

Consider:
$y = 5$
and
$y = x + ln(x)$

Take an educated guess at a solution.

$y_n = x_n + ln(x_n)$
_
Consider the slope of the tangent line to this function at $x_n$.

The slope of the tangent line of this function at $x_n$ is approximately equal to the slope of the secant line of this function between $x_n$ and $x_{n+1}$.

Let $y_{n+1} = 5$.

Find your new $x_{n+1}$.

Repeat iteratively.

Within 3 iterations (by hand), I converged on the solution.

5. Jul 7, 2015

### HallsofIvy

Staff Emeritus
The "Lambert W function" that Orodruin referred to is defined as the inverse function to $f(x)= xe^x$. So, immediately, x= W(e^5).