Question

Solve: \[2{\log _4}\left( {4 - x} \right) = 4 - {\log _2}\left( { - 2 - x} \right)\].

Answer 1

Hint:
First, we will apply the log rule,\[{\log _a}b = \dfrac{{{{\log }_c}b}}{{{{\log }_c}a}}\] in right hand side of the above equation and simplify. Then we will add \[{\log _4}\left( { - 2 - x} \right)\] to both sides of the above equation and add the log rule, \[{\log _c}a + {\log _c}b = {\log _c}ab\] in the obtained equation. Then we will use the logarithmic definition, if \[{\log _a}b = c\], then \[b = {a^c}\] in the obtained equation and then simplify to find the required value.

Complete step by step solution:
We are given that \[2{\log _4}\left( {4 - x} \right) = 4 - {\log _2}\left( { - 2 - x} \right)\].
Applying the log rule,\[{\log _a}b = \dfrac{{{{\log }_c}b}}{{{{\log }_c}a}}\] in right hand side of the above equation and simplify, we get
\[
   \Rightarrow {\log _2}\left( { - 2 - x} \right) = 4 - \dfrac{{{{\log }_4}\left( { - 2 - x} \right)}}{{{{\log }_4}2}} \\
   \Rightarrow 2{\log _2}\left( { - 2 - x} \right) = 4 - 2{\log _4}\left( { - 2 - x} \right) \\
 \]
Dividing both sides by 2 in the above equation, we get
\[
   \Rightarrow \dfrac{2}{2}{\log _2}\left( { - 2 - x} \right) = \dfrac{4}{2} - \dfrac{{2{{\log }_4}\left( { - 2 - x} \right)}}{2} \\
   \Rightarrow {\log _2}\left( { - 2 - x} \right) = 2 - {\log _4}\left( { - 2 - x} \right) \\
 \]
Adding \[{\log _4}\left( { - 2 - x} \right)\] to both sides of the above equation, we get
\[
   \Rightarrow {\log _2}\left( { - 2 - x} \right) + {\log _4}\left( { - 2 - x} \right) = 2 - {\log _4}\left( { - 2 - x} \right) + {\log _4}\left( { - 2 - x} \right) \\
   \Rightarrow {\log _2}\left( { - 2 - x} \right) + {\log _4}\left( { - 2 - x} \right) = 2 \\
 \]
Applying the log rule, \[{\log _c}a + {\log _c}b = {\log _c}ab\] in the above equation, we get
\[ \Rightarrow {\log _4}\left( {\left( {4 - x} \right)\left( { - 2 - x} \right)} \right) = 2\]
Using the logarithmic definition, if \[{\log _a}b = c\], then \[b = {a^c}\] in the above equation, we get
\[
   \Rightarrow \left( {4 - x} \right)\left( { - 2 - x} \right) = {4^2} \\
   \Rightarrow \left( {4 - x} \right)\left( { - 2 - x} \right) = 16 \\
   \Rightarrow - 8 - 4x + 2x + {x^2} = 16 \\
   \Rightarrow - 8 - 2x + {x^2} = 16 \\
 \]
Subtracting the above equation by 16 on both sides, we get
\[
   \Rightarrow - 8 - 2x + {x^2} - 16 = 16 - 16 \\
   \Rightarrow - 24 - 2x + {x^2} = 0 \\
   \Rightarrow {x^2} - 2x - 24 = 0 \\
 \]
Factorizing the above equation by splitting the middle term, we get
\[
   \Rightarrow {x^2} - 6x + 4x - 24 = 0 \\
   \Rightarrow x\left( {x - 6} \right) + 4\left( {x - 6} \right) = 0 \\
   \Rightarrow \left( {x + 4} \right)\left( {x - 6} \right) = 0 \\
 \]
Taking \[x + 4 = 0\] and simplifying it, we get
\[ \Rightarrow x = - 4\]
Taking \[x - 6 = 0\] and simplifying it, we get
\[ \Rightarrow x = 6\]
But since we know that \[x\] in the function \[\log x\] is always greater than 0 and when \[x = 6\], we have
\[ \Rightarrow - 2 - 6 = - 8\]

So, the value of \[x\] is \[ - 4\].

Note:
The key point here is to use the properties of the logarithm and the trigonometric rule right time in the question or else it will be really confusing to solve. 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 \[e\].