Question

If the logarithmic expression $\sqrt{{{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{{{x}^{2}}-1} \right)}$ is real, then determine the value of $'x'$ for which this will hold.

Answer 1

We solve this problem by checking where the given expression will be real. We use the condition that if an expression of the form $\sqrt{x}$ is real if and only if $x > 0$ then we solve the given expression by using the above condition to get the domain of $'x'$ where the given expression is real.
We use the condition of inequalities that is if
${{\log }_{a}}b > k$
Then depending on value of $'a'$ we have two conditions that is
$\Rightarrow b > {{a}^{k}}\left( \forall a\ge 1 \right)$
$\Rightarrow b < {{a}^{k}}\left( \forall 0 < a < 1 \right)$

Complete step-by-step solution
We are given that the expression as $\sqrt{{{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{{{x}^{2}}-1} \right)}$
We are given that this expression is real
We know that if an expression of the form $\sqrt{x}$ is real if and only if $x > 0$
By using the above condition to given expression we get
$\Rightarrow {{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{{{x}^{2}}-1} \right) > 0$
We know that the condition of logarithms that is if
${{\log }_{a}}b > k$
Then depending on value of $'a'$ we have two conditions that is
$\Rightarrow b > {{a}^{k}}\left( \forall a\ge 1 \right)$
$\Rightarrow b < {{a}^{k}}\left( \forall 0 < a < 1 \right)$
By using the above condition we get

& \Rightarrow \dfrac{x}{{{x}^{2}}-1} < {{\left( \dfrac{1}{2} \right)}^{0}} \\\ & \Rightarrow \dfrac{x}{{{x}^{2}}-1} < 1 \\\ \end{aligned}$$ Now, by taking the terms on one side we get $$\begin{aligned} & \Rightarrow \dfrac{x}{{{x}^{2}}-1}-1 < 0 \\\ & \Rightarrow \dfrac{-\left( {{x}^{2}}-x-1 \right)}{\left( x-1 \right)\left( x+1 \right)} < 0 \\\ \end{aligned}$$ By taking the negative sign to other side of inequality we know that the inequality changes then we get $$\Rightarrow \dfrac{\left( {{x}^{2}}-x-1 \right)}{\left( x-1 \right)\left( x+1 \right)} > 0.......equation(i)$$ Now let us take the quadratic equation of the numerator we get $$\Rightarrow f\left( x \right)={{x}^{2}}-x-1$$ We know that the zeros of quadratic equation $$a{{x}^{2}}+bx+c$$ are given as $$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ By using the above formula we get the zeros of $$f\left( x \right)$$ are $$\begin{aligned} & \Rightarrow x=\dfrac{1\pm \sqrt{1+4}}{2} \\\ & \Rightarrow x=\dfrac{1\pm \sqrt{5}}{2} \\\ \end{aligned}$$ We know that if $$\alpha ,\beta $$ are zeros of $$a{{x}^{2}}+bx+c$$ then we can write the expression as $$\Rightarrow a{{x}^{2}}+bx+c=\left( x-\alpha \right)\left( x-\beta \right)$$ By using this condition we get $$\Rightarrow f\left( x \right)=\left( x-\dfrac{1+\sqrt{5}}{2} \right)\left( x-\dfrac{1-\sqrt{5}}{2} \right)$$ By substituting the value in equation (i) we get $$\Rightarrow \dfrac{\left( x-\dfrac{1+\sqrt{5}}{2} \right)\left( x-\dfrac{1-\sqrt{5}}{2} \right)}{\left( x-1 \right)\left( x+1 \right)} > 0$$ Now, by drawing the number line and using the wave theory we get  Here, we can see that the range on the number line where it was positive is the required answer. Therefore, we can conclude that the domain of $$'x'$$ for which the given expression is real is $$\therefore x\in \left( -1,\dfrac{1-\sqrt{5}}{2} \right)\cup \left( 1,\dfrac{1+\sqrt{5}}{2} \right)$$ **Note:** Students may make mistakes in inequalities of logarithms. We have the condition that is if $${{\log }_{a}}b > k$$ Then depending on value of $$'a'$$ we have two conditions that is $$\Rightarrow b > {{a}^{k}}\left( \forall a\ge 1 \right)$$ $$\Rightarrow b < {{a}^{k}}\left( \forall 0 < a < 1 \right)$$ But students may consider the same result in both cases as $$\Rightarrow b > {{a}^{k}}\left( \forall a \right)$$ This may give the wrong answer.