Question

Find the value of \[2{\log _4}\left( {4 - x} \right) = 4 - {\log _2}\left( { - 2 - x} \right)\].

Answer 1

Hint: Here, we will start by converting the \[{\log _4}\] to \[{\log _2}\] term. We can do this by using the change-of-base formula, \[{\log _a}x = \dfrac{{{{\log }_b}x}}{{{{\log }_b}a}}\] in the above equation and then use the logarithm property, \[{\log _b}\dfrac{a}{c} = {\log _b}a - {\log _b}c\]and then the power rule of logarithm, \[{\log _b}\left( {{a^c}} \right) = c{\log _b}a\] to simplify the equations to find the required value.

Complete step by step answer:

We are given equation
\[2{\log _4}\left( {4 - x} \right) = 4 - {\log _2}\left( { - 2 - x} \right){\text{ ......eq.(1)}}\]
We know that the domain of \[{\log _4}\left( {4 - x} \right)\] is \[4 - x > 0\] and domain of \[{\log _2}\left( { - 2 - x} \right)\] is \[ - 2 - x\].
Therefore, the domain is \[x \in \left( { - \infty , - 2} \right)\].
Let us now start by converting the \[{\log _4}\] to \[{\log _2}\] term. We can do this by using the change-of-base formula, \[{\log _a}x = \dfrac{{{{\log }_b}x}}{{{{\log }_b}a}}\] in the above equation, we get
\[ \Rightarrow 2\dfrac{{{{\log }_2}\left( {4 - x} \right)}}{{{{\log }_2}2}} = 4 - {\log _2}\left( { - 2 - x} \right)\]
Using the logarithm property \[{\log _a}a = a\], in the above equation, we get
\[
   \Rightarrow {\log _2}\left( {4 - x} \right) = 4 - {\log _2}\left( { - 2 - x} \right) \\
   \Rightarrow {\log _2}\left( {4 - x} \right) = \log {e^4} - {\log _2}\left( { - 2 - x} \right) \\
 \]
Using the logarithm property, \[{\log _b}\dfrac{a}{c} = {\log _b}a - {\log _b}c\] in the above expression, we get
\[ \Rightarrow {\log _2}\left( {4 - x} \right) = {\log _2}\left( {\dfrac{{{e^4}}}{{ - 2 - x}}} \right)\]
Let us now make use of the power rule of logarithm, \[{\log _b}\left( {{a^c}} \right) = c{\log _b}a\].
So, on applying this rule in the left-hand side in the above equation, we get
\[ \Rightarrow {\log _2}\left( {4 - x} \right) = {\log _2}\left( {\dfrac{{{e^4}}}{{ - 2 - x}}} \right)\]
We know if the logs are equal, then their argument must be equal in the above equation, we get
\[ \Rightarrow \left( {4 - x} \right) = \dfrac{{16}}{{ - 2 - x}}\]
Cross-multiplying the above equation, we get
\[
   \Rightarrow \left( {x - 4} \right)\left( { - 2 - x} \right) = 16 \\
   \Rightarrow {x^2} - 2x - 8 = 16 \\
 \]

Factorizing the above equation, we get
\[
   \Rightarrow {x^2} - 6x + 4x - 24 = 0 \\
   \Rightarrow \left( {x - 6} \right)\left( {x + 4} \right) = 0 \\
   \Rightarrow x - 6 = 0{\text{ or }}x + 4 = 0 \\
   \Rightarrow x = 6, - 4 \\
 \]
As \[x = 6\] is not in the domain,\[x = - 4\] is the solution.

Note: The power rule can be used for fast exponent calculation using multiplication operation. Students should make use of the appropriate formula of logarithms wherever needed and solve the problem. In mathematics, if the base value in the logarithm function is not written, then the base is \[e\].