Question

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

Answer 1

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

& \Rightarrow 4-x> 0 \\\ & \Rightarrow x<4........equation(i) \\\ \end{aligned}$$ Now, by taking the second logarithm from the equation we get $$\begin{aligned} & \Rightarrow -2-x >0 \\\ & \Rightarrow x< -2.........equation(ii) \\\ \end{aligned}$$ Now, by combining the equation (i) and equation (ii) we get $$\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{aligned} & \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{aligned}$$ Now, we know that the sum of logarithms formula that is $${{\log }_{b}}a+{{\log }_{b}}c={{\log }_{b}}\left( a\times c \right)$$ By using the this formula to above equation we get $$\begin{aligned} & \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{aligned}$$ We know that the definition of a logarithm that is if the equation of form $${{\log }_{b}}a=N$$ then $$a={{b}^{N}}$$ By using the this definition to above equation we get $$\begin{aligned} & \Rightarrow {{x}^{2}}-2x-8={{2}^{4}} \\\ & \Rightarrow {{x}^{2}}-2x-24=0 \\\ \end{aligned}$$ We now that the formula for roots of quadratic equation $$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{aligned} & \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{aligned}$$ Now, by taking the number 4 outside of square root we get $$\begin{aligned} & \Rightarrow x=\dfrac{2\pm 2\sqrt{25}}{2} \\\ & \Rightarrow x=1\pm 5 \\\ \end{aligned}$$ Here, we can see that we get two values of $$'x'$$ one for positive sign and one for negative sign. So, we get the values of $$'x'$$ as $$\begin{aligned} & \Rightarrow x=\left( 1+5 \right),\left( 1-5 \right) \\\ & \Rightarrow x=6,-4 \\\ \end{aligned}$$ But we have the domain of $$'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{aligned} & \Rightarrow x=\left( 1+5 \right),\left( 1-5 \right) \\\ & \Rightarrow x=6,-4 \\\ \end{aligned}$$ We need to check whether the function can hold the roots we got or not. We know that the logarithm $${{\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.