Question

Solve the following equation:
\[2{{\log }_{4}}\left( 4-x \right)=4-{{\log }_{2}}\left( -2-x \right)\]

Answer 1

Hint: We solve this problem by using the conditions of logarithms.
(1) The logarithm is of form \[{{\log }_{b}}a\] exists if and only if \[a,b > 0\]
(2) If there is an equation that is \[{{\log }_{b}}a={{\log }_{b}}c\] then we can say that \[a=c\]
(3) We have a standard formula of logarithms that is
\[{{\log }_{{{b}^{c}}}}a=\dfrac{1}{c}{{\log }_{b}}a\]
\[{{\log }_{b}}a+{{\log }_{b}}c={{\log }_{b}}\left( a\times c \right)\]
(4) If the equation of form \[{{\log }_{b}}a=N\] then \[a={{b}^{N}}\]
By using the above formulas we calculate the value of \['x'\] that satisfies the given equation.

Complete step-by-step solution
We are given that the equation as
\[2{{\log }_{4}}\left( 4-x \right)=4-{{\log }_{2}}\left( -2-x \right)\]
We know that \[{{\log }_{b}}a\] exists if and only if \[a,b >0\]
By using the above condition and taking the first logarithm we get
\[\begin{align}
  & \Rightarrow 4-x> 0 \\
 & \Rightarrow x<4........equation(i) \\
\end{align}\]
Now, by taking the second logarithm from the equation we get
\[\begin{align}
  & \Rightarrow -2-x >0 \\
 & \Rightarrow x< -2.........equation(ii) \\
\end{align}\]
Now, by combining the equation (i) and equation (ii) we get
\[\Rightarrow x< -2\]
Here, by converting the above equation in to domain we get
\[\Rightarrow x\in \left( -\infty ,-2 \right).........equation(iii)\]
Now, by the given equation and taking all the logarithm terms to one side we get
\[\Rightarrow 2{{\log }_{4}}\left( 4-x \right)+{{\log }_{2}}\left( -2-x \right)=4\]
We know that the formula of logarithms that is
\[{{\log }_{{{b}^{c}}}}a=\dfrac{1}{c}{{\log }_{b}}a\]
By using the above formula we get
\[\begin{align}
  & \Rightarrow \dfrac{2}{2}{{\log }_{2}}\left( 4-x \right)+{{\log }_{2}}\left( -2-x \right)=4 \\
 & \Rightarrow {{\log }_{2}}\left( 4-x \right)+{{\log }_{2}}\left( -2-x \right)=4 \\
\end{align}\]
Now, we know that the sum of logarithms formula that is
\[{{\log }_{b}}a+{{\log }_{b}}c={{\log }_{b}}\left( a\times c \right)\]
By using the this formula to above equation we get
\[\begin{align}
  & \Rightarrow {{\log }_{2}}\left[ \left( 4-x \right)\left( -x-2 \right) \right]=4 \\
 & \Rightarrow {{\log }_{2}}\left[ {{x}^{2}}-2x-8 \right]=4 \\
\end{align}\]
We know that the definition of a logarithm that is if the equation of form \[{{\log }_{b}}a=N\] then \[a={{b}^{N}}\]
By using the this definition to above equation we get
\[\begin{align}
  & \Rightarrow {{x}^{2}}-2x-8={{2}^{4}} \\
 & \Rightarrow {{x}^{2}}-2x-24=0 \\
\end{align}\]
We now that the formula for roots of quadratic equation \[a{{x}^{2}}+bx+c=0\] is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -24 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{2\pm \sqrt{4\left( 1+24 \right)}}{2} \\
\end{align}\]
Now, by taking the number 4 outside of square root we get
\[\begin{align}
  & \Rightarrow x=\dfrac{2\pm 2\sqrt{25}}{2} \\
 & \Rightarrow x=1\pm 5 \\
\end{align}\]
Here, we can see that we get two values of \['x'\] one for positive sign and one for negative sign.
So, we get the values of \['x'\] as
\[\begin{align}
  & \Rightarrow x=\left( 1+5 \right),\left( 1-5 \right) \\
 & \Rightarrow x=6,-4 \\
\end{align}\]
But we have the domain of \['x'\] from equation (iii) that is
\[\Rightarrow x\in \left( -\infty ,-2 \right)\]
We know that the values of \['x'\] that lie in the domain \[\left( -\infty ,-2 \right)\] satisfy the given equation.
But we get the one of the roots of equation as
\[\Rightarrow x=6>-2\]
So, we can say that the above root doesn’t satisfy the given equation.
Therefore, the solution of given equation is \[x=-4\]

Note: Students may do mistake in the roots of given equation.
We got the roots of equation as
\[\begin{align}
& \Rightarrow x=\left( 1+5 \right),\left( 1-5 \right) \\
& \Rightarrow x=6,-4 \\
\end{align}\]
We need to check whether the function can hold the roots we got or not.
We know that the logarithm \[{{\log }_{b}}a\] exists if and only if \[a,b>0\]
Students may this point and gives the both values as roots of given equation which is wrong answer.