Question

How do you solve $ {\log _4}\left( {{x^2} - 4} \right) - {\log _4}\left( {x + 2} \right) = 2 $ ?

Answer 1

Hint: To solve the problem we should know about the following term.
Logarithm: The exponent that indicates the power to which a base number is raised to produce a given number. the logarithm of $ 100 $ to the base $ 10 $ is $ 2 $

Complete step by step solution:
As given in the quest
  $ {\log _4}\left( {{x^2} - 4} \right) - {\log _4}\left( {x + 2} \right) = 2 $
Let’s determine the set of real numbers where this equation makes sense. It’s necessary because we might engage in non- invariant transformation of this equation, which might produce extra solutions or we might lose certain solutions in the course of transformation. We will use only invariant transformations. It’s still a good practice to determine what kind of solutions our equation allows.
Any logarithm with a base $ 4 $ (any other positive base) is defined only for positive arguments. That why we have restrictions:
  $ {x^2} - 4 > 0 $
  $ x + 2 > 0 $
The first condition is equivalent to $ {x^2} > 4 $ or $ \left| x \right| > 2 $ a combination of two inequalities:
  $ x < - 2 $ or,
  $ x > 2 $
The second condition is equivalent to
  $ x > - 2 $
The only condition that satisfies condition (3) and either (1) or (2) above is
  $ x > 2 $
Now let’s examine the problem at hand.
Recall that
  $ {\log _c}(a \times b) = {\log _c}(a) + {\log _c}(b) $
Applying this to our equation and using an identify $ {x^2} - 4 = (x - 2) \times (x + 2) $ , we can rewrite it as $ \log $
  $ {\log _4}\left( {x - 2} \right) + {\log _4}\left( {x + 2} \right) - {\log _4}(x + 2) = 2 $
Or,
  $ \Rightarrow {\log _4}(x - 2) = 2 $
  $ \Rightarrow {4^2} = x - 2 $ or,
  $ \Rightarrow 16 = x - 2 $ or,
  $ \Rightarrow x = 18 $
Hence, $ x = 18 $
So, the correct answer is “ $ x = 18 $ ”.

Note: More generally, exponentiation allow any positive real number as base to be raised to any real power, always producing a positive result, so $ {\log _b}\left( x \right) $ for any two positive real numbers $ b $ and $ x $ where $ b $ is not equal to $ 1 $ , is always a unique real number $ y $ .