TITLE: Solving $\ln(x^2+1)+1 = \ln(x^2+4)$
QUESTION [6 upvotes]: This is a homework question, but I've tried as hard as I can. Let me walk you through what I've done so far.
$$\ln(x^2+1)+1 = \ln(x^2+4)$$
$$\ln(x^2+4) - \ln(x^2+1) = 1$$
$$\ln\left(\frac{x^2+4}{x^2+1}\right) = 1$$
Now, this is where I'm kind of getting lost. Maybe I should rewrite the equation? Doesn't this basically say “e to the 1st power should be equal to $\frac{x^2+4}{x^2+1}$”?
$$\frac{x^2+4}{x^2+1} = e$$
$$x^2+4 = ex^2+e$$
This is where I couldn't move on, but as I was writing this post, this hit me:
$$x^2-ex^2 = e-4$$
$$(1-e)x^2 = e-4$$
$$x^2 = \frac{e-4}{1-e}$$
$$x = \pm \sqrt{\frac{e-4}{1-e}}$$
According to my book, the answer should be:
$$x = \pm \sqrt{\frac{4-e}{e-1}}$$
By calculating the right-hand expression, I see that it is the same as my answer.
$$x \approx \pm 0.86$$
Two questions:

What's the reason for changing the order of terms in the solution?
Have I made any particularly odd steps in my solution?

REPLY [4 votes]: Remember that $$\dfrac{y}{z} = \dfrac{-y}{-z}$$
Hence,
$$\dfrac{e-4}{1-e} = \dfrac{4-e}{e-1}$$ A possible reason why the book gives the answer as $\dfrac{4-e}{e-1}$ is that $4-e$ and $e-1$ are both positive. It is generally preferred to write a positive fraction as a ratio of two positive numbers as opposed to two negative numbers.