Solve ${\log _2}x + {\log _2}\left( {x - 2} \right) = 3$.

Hint:We know the product rule: ${\log _b}\left( P \right) + {\log _b}\left( Q \right) = {\log _b}\left( {PQ} \right)$
So using the product rule we can simplify the above logarithmic terms.Also we know:
Quadratic Formula: $a{x^2} + bx + c = 0$ Here $a,\;b,\;c$ are numerical coefficients.
So to solve $x$ we have:$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So by using the above formulas and identities we can simplify and thus solve the given question.

Complete step by step answer:
Given, ${\log _2}x + {\log _2}\left( {x - 2} \right) = 3.......................\left( i \right)$
So in order to solve equation (i) we first have to simplify the given equation and then solve for $x$ using quadratic formula. On observing (i) we can say that the identity ${\log _b}\left( P \right) + {\log _b}\left( Q \right) = {\log _b}\left( {PQ} \right)$ can be used in the terms ${\log _2}x + {\log _2}\left( {x - 2} \right)$. Such that we can write:
${\log _2}x + {\log _2}\left( {x - 2} \right) = {\log _2}\left( {x\left( {x - 2} \right)} \right)...................\left( {ii} \right)$
Substituting (ii) in (i) we can write:
${\log _2}\left( {x\left( {x - 2} \right)} \right) = 3.....................\left( {iii} \right)$
Now we have the identity where we can convert the logarithmic equation to exponential equation by:
${\log _b}\left( P \right) = Q\;\; \Rightarrow P = {b^Q}$
Here on comparing we can write:
$b = 2 \\
\Rightarrow P = \left( {x\left( {x - 2} \right)} \right) \\
\Rightarrow Q = 3 \\ $
Such that we can write equation (iii) as:
\[{\log _2}\left( {x\left( {x - 2} \right)} \right) = 3 \\
\Rightarrow x\left( {x - 2} \right) = {2^3}....................\left( {iv} \right) \\ \]
Now we need to simplify (iii) such that we can apply the quadratic formula:
Such that:
\[x\left( {x - 2} \right) = {2^3} \\
\Rightarrow{x^2} - 2x = 8 \\
\Rightarrow{x^2} - 2x - 8 = 0....................\left( v \right) \\ \]
Now in (v) we can apply the quadratic formula since it’s in the form$a{x^2} + bx + c = 0$.
 So to solve $x$ we have:$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So on comparing (i) to the general formula we get:
$a = 1,\;b = - 2,\;c = - 8.....................\left( {vi} \right)$
Now substituting in the formula we can write:
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\Rightarrow x = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - \left( {4 \times 1 \times - 8} \right)} }}{{2 \times 1}} \\
\Rightarrow x = \dfrac{{2 \pm \sqrt {4 - \left( { - 32} \right)} }}{2} \\
\Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 32} }}{2} \\
\Rightarrow x = \dfrac{{2 \pm \sqrt {36} }}{2} \\
\Rightarrow x = \dfrac{{2 \pm 6}}{2} \\ \]
Now there are two possibilities of$x$, which is produced either by addition or by subtraction. It’s found such that:
\[x = \dfrac{{2 + 6}}{2}\;\;\;\;\;{\text{and}}\;\;\;x = \dfrac{{2 - 6}}{2} \\
\Rightarrow x = \dfrac{8}{2}\;\;\;\;\;\;\;\;{\text{and}}\;\;\;x = - \dfrac{4}{2} \\
\Rightarrow x = 4\;\;\;\;\;\;\;\;\;\;{\text{and}}\;\;\;x = - 2 \\
\therefore x = 4,\; - 2 \\ \]
Therefore on solving ${\log _2}x + {\log _2}\left( {x - 2} \right) = 3$ we get\[x = 4,\; - 2\].

Note:Some of the logarithmic properties useful for solving questions of derivatives are listed below:
$1.\;{\log _b}\left( {PQ} \right) = {\log _b}\left( P \right) + {\log _b}\left( Q \right) \\
2.\;{\log _b}\left( {\dfrac{P}{Q}} \right) = {\log _b}\left( P \right) - {\log _b}\left( Q \right) \\
3.\;{\log _b}\left( {{P^q}} \right) = q \times {\log _b}\left( P \right)$
These are called Product Rule, Quotient Rule and Power Rule respectively.Quadratic formula is mainly used in conditions where grouping method cannot be used or when the polynomial cannot be reduced into some general identity.Quadratic formula method is an easier and direct method on comparison to other methods.