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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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2  
(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  
+1 for showing what you've tried. –  Rick Decker Sep 20 '12 at 16:10

3 Answers 3

up vote 4 down vote accepted

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.

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

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You have not made any particularly odd steps in your solution. As you can see, your answer and the answer in the book are equal: $$\sqrt{\frac{e - 4}{1 - e}} = \sqrt{\frac{-(4 - e)}{-(e - 1)}} = \sqrt{\frac{4 - e}{e - 1}}.$$ At some point you arrive at the equation $x^2 + 4 = ex^2 + e$. In order to solve this equation, you put all the terms involving $x$ on one side and the constant terms on the other; here you have two choices: $x$ terms on the left, or $x$ terms on the right. Putting the $x$ terms on the left gives your final expression for $x$, while putting the $x$ terms on the right gives the book's expression.

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