# Solving $\ln(x^2+1)+1 = \ln(x^2+4)$

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:

1. What's the reason for changing the order of terms in the solution?
2. Have I made any particularly odd steps in my solution?
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(1) They apparently want the numerator and denominator to both be positive. If both are negative then it doesn't really matter (the negatives cancel in the fraction), so this is essentially a matter of taste, where someone may not like tacit negatives under a square root symbol. (2) I don't see any. – anon Sep 20 '12 at 14:43
A comment irrelevant to your question. Your initial equation is nice, and you manipulated it into something less nice, with negatives, division. Don't manipulate, exponentiate, as in the answer by Kevin. – André Nicolas Sep 20 '12 at 15:02
+1 for showing what you've tried. – Rick Decker Sep 20 '12 at 16:10
You have not made any particularly odd steps in your solution. As the answers below illustrate, your answer and the answer in the book are equal. In order to solve the equation $x^2 + 4 = ex^2 + e$, you put all the terms involving $x^2$ on one side and the constant terms on the other. Here you have two choices: $x^2$ terms on the left, or $x^2$ terms on the right. Putting the $x^2$ terms on the left gives your final expression for $x$, while putting the $x^2$ terms on the right gives the book's expression. – Michael Albanese Jun 21 '15 at 0:20

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.
Note that $\frac{(e-4)(-1)}{(1-e)(-1)}=\frac{4-e}{e-1}$