# What is x if log_2(3-x) + log_2 (2-x) = log_2 (1-x)?

Nov 9, 2015

No solution in $\mathbb{R}$.
Solutions in $\mathbb{C}$: $\textcolor{w h i t e}{\times x} 2 + i \textcolor{w h i t e}{\times x} \text{and} \textcolor{w h i t e}{\times x} 2 - i$

#### Explanation:

First, use the logarithm rule:

${\log}_{a} \left(x\right) + {\log}_{a} \left(y\right) = {\log}_{a} \left(x \cdot y\right)$

Here, this means that you can transform your equation as follows:

${\log}_{2} \left(3 - x\right) + {\log}_{2} \left(2 - x\right) = {\log}_{2} \left(1 - x\right)$
$\iff {\log}_{2} \left(\left(3 - x\right) \left(2 - x\right)\right) = {\log}_{2} \left(1 - x\right)$

At this point, as your logarithm basis is $> 1$, you can "drop" the logarithm on both sides since $\log x = \log y \iff x = y$ for $x , y > 0$.
Please beware that you can't do such a thing when there is still a sum of logarithms like in the beginning.

So, now you have:
${\log}_{2} \left(\left(3 - x\right) \left(2 - x\right)\right) = {\log}_{2} \left(1 - x\right)$
$\iff \left(3 - x\right) \left(2 - x\right) = 1 - x$
$\iff 6 - 5 x + {x}^{2} = 1 - x$
$\iff 5 - 4 x + {x}^{2} = 0$

This is a regular quadratic equation which you can solve in several different ways.

This one sadly doesn't have a solution for real numbers.

$\textcolor{B l u e}{\text{~~~~~~~~~~~~~~ proposed addition~~~~~~~~~~~~~~~~~}}$

Tony B:
$\textcolor{b l u e}{\text{I agree with your calculations and think they are well presented}}$

$\textcolor{b r o w n}{\text{if I may I would like to expand on your answer a little!}}$

I totally concur that there is no solution for $x \ne \mathbb{R}$
If on the other hand we look at the potential of $x \in \mathbb{C}$ then we are able to ascertain two solutions.

Using standard form
$a {x}^{2} + b c + c = 0 \textcolor{w h i t e}{\times \times} \text{where}$

$x = \frac{- b \pm \sqrt{{\left(- b\right)}^{2} - 4 a c}}{2 a}$

We then we end up with:

$\frac{+ 4 \pm 2 i}{2} \to \textcolor{w h i t e}{\times x} 2 + i \textcolor{w h i t e}{\times x} \text{and} \textcolor{w h i t e}{\times x} 2 - i$

Nov 10, 2015

My understanding implies that the question given needs to be checked. $\textcolor{b r o w n}{\text{If "x in RR" then it is indeterminate. On the other hand if "x notin RR" then this may not be the case.}}$

#### Explanation:

Pre-amble

Log addition is the consequence of multiplication of the source numbers/variables.

The equals sign is a $\textcolor{b l u e}{\text{mathematical}}$ absolute, stating that what is one side of its has the exact same intrinsic value that is on the other side.

Both sides of the equals sign are to log base 2. Suppose we had some random value of say $t$. If we had ${\log}_{2} \left(t\right) \text{ then antilog} \left[{\log}_{2} \left(t\right)\right] = t$ This type of mathematical notation is sometimes written as ${\log}_{2}^{-} 1 \left(t\right) = t$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Solution to this problem:

Take antilogs of both sides giving in the question implies:
$\left(3 - x\right) \left(2 - x\right) \to \left(1 - x\right)$

This I believe to be $\textcolor{red}{\text{indeterminate}}$ in that the LHS does not have exactly the same intrinsic value as the RHS. This$\textcolor{g r e e n}{\text{ implies}}$ that the question may need to be worded differently.

$\textcolor{b r o w n}{\text{On the other hand it may be the case that } x \in \mathbb{C}}$.
$\textcolor{b r o w n}{\text{This may well produce an answer.}}$

$\left(3 - x\right) \left(2 - x\right) = {x}^{2} - 5 x + 6 \ne \left(1 - x\right) \text{ for } x \in \mathbb{R}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\left(3 - x\right) \left(2 - x\right) = {x}^{2} - 5 x + 6 = \left(1 - x\right) \text{ for } x \in \mathbb{C}$

x = 2+i ; 2-i