Question

Solve the expression ${{\log }_{0.2}}{{\log }_{6}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)=0$ for x.

Answer 1

Hint: We will use the logarithm formula which comes along with exponential. The formula is numerically understood as ${{\log }_{e}}=x$ converted into $1={{e}^{x}}$ where e is called the exponential and log is the logarithm. With the help of this formula we will be able to solve the question. We will also use cross multiplication here which is done as method $\dfrac{a}{b}=\dfrac{c}{d}$ resulting into ad = cb.

Complete step-by-step answer:
Now we will first consider the expression ${{\log }_{0.2}}{{\log }_{6}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)=0...(i)$. Now we will use the formula ${{\log }_{e}}=x$ converted into $1={{e}^{x}}$ which results into a new equation written as $\begin{aligned} & {{\log }_{0.2}}{{\log }_{6}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)=0 \\\ & \Rightarrow {{\log }_{6}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)={{\left( 0.2 \right)}^{0}} \\\ \end{aligned}$
As we know that any number with the power as 0 will result only into 1. Therefore, the equation ${{\log }_{6}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)={{\left( 0.2 \right)}^{0}}$ is converted into ${{\log }_{6}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)=1$.
Now we will again apply the formula ${{\log }_{e}}=x$ converted into $1={{e}^{x}}$. Therefore, we get
$\begin{aligned} & {{\log }_{6}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)=1 \\\ & \Rightarrow \left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)={{\left( 6 \right)}^{1}} \\\ & \Rightarrow \left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)=6 \\\ \end{aligned}$
Now we will do cross multiplication here which is done by using the method $\dfrac{a}{b}=\dfrac{c}{d}$ resulting into ad = cb. Therefore we will now have
$\begin{aligned} & \left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)=6 \\\ & \Rightarrow {{x}^{2}}-1=6\left( {{x}^{2}}+1 \right) \\\ & \Rightarrow {{x}^{2}}-1=6{{x}^{2}}+6 \\\ \end{aligned}$
Now we will take all the terms to the right of equal sign. Therefore we get
$\begin{aligned} & {{x}^{2}}-1=6{{x}^{2}}+6 \\\ & \Rightarrow 6{{x}^{2}}+6-{{x}^{2}}+1=0 \\\ \end{aligned}$
After doing simplification we will have
$\begin{aligned} & 6{{x}^{2}}+6-{{x}^{2}}+1=0 \\\ & \Rightarrow 5{{x}^{2}}+7=0 \\\ \end{aligned}$
Now we will take 7 to the right side of the equal sign. Thus we get $5{{x}^{2}}=-7$. After dividing the equation by 5 on both the sides we will have ${{x}^{2}}=-\dfrac{7}{5}$. Now we will apply the square root on both the sides of the equation. Thus, we get $\sqrt{{{x}^{2}}}=\sqrt{-\dfrac{7}{5}}$. This can also be written as $x=\sqrt{-1\times \dfrac{7}{5}}$. As we know that $\sqrt{-1}=i$. Thus, we get $x=i\sqrt{\dfrac{7}{5}}$.

Note: When we get a negative sign inside the root we start thinking that our answer is incorrect or we just stop there. But actually this is not the case here. Here the negative sign inside the root indicates the complex term which is represented by i. This is called iota. We can also solve the equation $5{{x}^{2}}+7=0$ by the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We can use this formula by substituting a = 5, b = 0 and c = 7. And we will get the desired answer. Instead of using cross multiplication we can also solve $\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)=6$ we can also multiply ${{x}^{2}}+1$ to both sides of the equation. By this also we will get the required answer.