The problem is:

Given this equation find the possible values of $x$. $$\log_{2}\left ( \frac{x+2}{x-1} \right )+\log_{2}\left [ (x+2)(x-1) \right ]=2$$

First I defined the domain of the function.

  1. The denominator $x-1$ must not be zero: $x-1\neq0$.

  2. The fraction $\frac{x+2}{x-1}$ must be bigger than zero: $\frac{x+2}{x-1}>0$.

  3. The product $(x+2)(x-1)$ must be bigger than zero: $(x+2)(x-1)>0$.

In the end I concluded that the domain should be $\left ]-\infty,-2\right [\cup \left ]1,+\infty\right[$.

Then I sarted to solve:

$$\begin{align*} \log_{2}\left ( \frac{x+2}{x-1} \right )&+\log_{2}\left [ (x+2)(x-1) \right ]=2 \\ &\iff \log_{2}(x+2)-\log_{2}(x-1)+\log_{2}(x+2)+\log_{2}(x-1)=2\\ &\iff 2 \cdot \log_{2}(x+2)=2\\ &\iff \log_{2}(x+2)=1 \\ &\iff x+2=2^1\\ &\iff x=2-2\\ &\iff x=0. \end{align*}$$

This result is not in the domain! Where I miss?

  • $\begingroup$ Put every step into wolfram alpha and look where it becomes wrong, learn to help yourself :-) $\endgroup$
    – Listing
    Oct 29, 2011 at 22:41
  • $\begingroup$ So the answer is: for no (real) value of $x$. $\endgroup$ Oct 29, 2011 at 22:43
  • 1
    $\begingroup$ I took the liberty of putting some periods to your text. This may sound rude, but it's for your own good: You wouldn't write an entire paragraph of text with no periods, would you? It's better that your learn to write math such that it flows like a "normal" text now, rather than when you hand in an assignment and get "add more periods and commas" as a comment from your teacher. $\endgroup$
    – kahen
    Oct 29, 2011 at 22:50
  • $\begingroup$ @Yuval: The first "if and only if" does not hold over the entire domain, only on $(1,\infty)$. $\endgroup$ Oct 29, 2011 at 22:54

2 Answers 2


One problem is that you can have both the product and the fraction be valid arguments for $\log_2$, but the individual factors/denominator/numerator not be valid arguments. Specifically, if $x\lt -2$, then neither $x-1$ nor $x+2$ are valid arguments, so you cannot take the first step in general.

Instead, why not work those identities in the other direction? $$\begin{align*} \log_2\left(\frac{x+2}{x-1}\right) + \log_2\Bigl( (x+2)(x-1)\Bigr) &= \log_2\left(\left( \frac{x+2}{x-1}\right)(x+2)(x-1)\right)\\ &= \log_2\left(\bigl(x+2\bigr)^2\right). \end{align*}$$ Now, if $x+2\gt0$ (and hence, if $x\gt 1$, then this is equal to $2\log_2(x+2)$, from which you conclude that $\log_2(x+2) = 1$, hence $x+2=2$, hence $x=0$, but that is not in the domain.

What does that mean? It means that you cannot find a solution with $x+2\gt 0$.

What if $x+2\lt 0$, though? Then simply go from $\log_2((x+2)^2) = 2$ to $(x+2)^2 = 2^2 = 4$, and hence $-(x+2) = 2$, or $x+2=-2$, hence $x=-4$. This is in the domain, and satisfies $x+2\lt 0$.

Indeed, note that if $x=-4$, then $\frac{x+2}{x-1} = \frac{-2}{-5} = \frac{2}{5}$, and $(x+2)(x-1) = (-2)(-5) = 10$, so you have that the left hand side is $$\log_2(2/5) + \log_2(10) = \log_2(20/5) = \log_2(4) = 2,$$ so $x=-4$ is a solution.


If $x>1$, then $\log_2\left(\dfrac{x+2}{x-1}\right) = \log_2(x+2)-\log_2(x-1)$.

If $x<-2$ then $\log_2\left(\dfrac{x+2}{x-1}\right) = \log_2\left(\dfrac{-x-2}{1-x}\right) = \log_2(-x-2)-\log_2(1-x)$.

A similar thing applies to the logarithm of the product.


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