# Losing valid logarithm solutions as a consequence of the order of solution steps?

Take, for example, the following:

$\ln(x-1)^2 = 4$

Given the rules of logarithms, we have two potential first steps. In one, we could move the exponent within the argument of the logarithm to be a coefficient:

\begin{aligned} \ln(x-1)^2 &= 4 \\ 2\ln(x-1) &= 4 \\ \ln(x-1) &= 2 \\ e^{\ln(x-1)} &= e^2 \\ x-1 &= e^2 \\ x &= 1+e^2 \\ \end{aligned}

Or, in another, we could first raise the base to each side, resulting, in this case, in:

\begin{aligned} \ln(x-1)^2 &= 4 \\ e^{\ln(x-1)^2} &= e^4 \\ (x-1)^2 &= e^4 \\ x-1 &= \pm e^2 \\ x &= 1 \pm e^2 \\ \end{aligned}

For each of these initial steps (as written, at least), the end results are slightly different. In the first case, carrying through to the end yields $\ x = 1 + e^2$, while the second yields two solutions: $\ x = 1 \pm e^2$.

The question is: How do we avoid missing one of the two solutions simply by virtue of our chosen order of solution steps? I.e., what am I missing here?

• The first approach is correct. The issue with the second solution is that for the second solution, $x = 1 - e^2$, this is not a solution since logarithms do not take arguments which are negative. EDIT: this comment is wrong, ignore it. Apr 18, 2018 at 19:45
• @Onetimething: The exponent is even, so it is fine (since the exponent is part of the argument of the logarithm). Apr 18, 2018 at 19:45
• sorry, missed that. Apr 18, 2018 at 19:47

You can avoid losing solutions if you take care of preserving the domain of the original equation, i.e. using $ln(x-1)^2=2\cdot ln|x-1|$

You have to take care when moving the exponent because $(x-1)^2=(1-x)^2$. In particular, solutions of $$\ln(x-1)^2=4$$ will be the same as the solutions to both $$2\ln(x-1)=4\quad\text{and}\quad2\ln(1-x)=4.$$This is true for any even exponent.

The issue is with

$\ln(x-1)^2 = 4$

$2\ln(x-1) = 2$

$log((-10)^2))= 2$
$2log(-10)=2$
$2\ln|x-1| = 2$