10 December 2021

Statement

Solve the following equation for \(x\).

\(\ln{x} + \ln{(x-1)} = 1\).

Solution

\(\ln{x} + \ln{(x-1)} = 1\)

We can use the property \(log_{c}{a} + log_{c}{b} = log_{c}{a \times b}\) and we get

\(\Leftrightarrow \ln{(x\times(x-1))} = 1\)

Since both sides are equal we can exponentiate and keep the equality

\(\Leftrightarrow e^{\ln{(x^2-x)}} = e\)

\(\Leftrightarrow x^2 - x = e\)

\(\Leftrightarrow x^2 - x - e = 0\)

The final step is to use the quadratic formula to find \(x\)

\(\Leftrightarrow x = \frac{1 \pm \sqrt{1 - 4 \times 1 \times (-e)}}{2}\)

\(\Leftrightarrow x = \frac{1 + \sqrt{1 + 4e}}{2} \lor x = \frac{1 - \sqrt{1 + 4e}}{2} \)

Notice that if we substitute \(x = \frac{1 - \sqrt{1 + 4e}}{2}\) in the original equation we get a logarithm with a negative argument, which is not defined, therefore the only real solution is

\(x = \frac{1 + \sqrt{1 + 4e}}{2}\)