Question

Find the characteristics of the solutions of the equation ${{\log }_{{{x}^{2}}}}16+{{\log }_{2x}}64=3$.
A. one irrational solution
B. no prime solution
C. two real solutions
D. one integral solutions

Answer 1

Hint: We need to first apply different logarithmic operations to simplify the given equation. We need to convert the complicated forms into their basic ones using ${{\log }_{{{a}^{n}}}}{{b}^{m}}=\dfrac{m}{n}{{\log }_{a}}b$ , ${{\log }_{a}}b=\dfrac{1}{{{\log }_{b}}a}$ . We get a quadratic equation after assuming the basic log value as variable. We solve the equation to find the solution of the problem.

Complete step by step answer:
We try to solve the given equation using different logarithmic relations.
$\begin{align}
  & {{\log }_{{{x}^{2}}}}16+{{\log }_{2x}}64=3 \\
 & \Rightarrow {{\log }_{{{x}^{2}}}}\left( {{2}^{4}} \right)+{{\log }_{2x}}\left( {{2}^{6}} \right)=3 \\
\end{align}$
We know that ${{\log }_{{{a}^{n}}}}{{b}^{m}}=\dfrac{m}{n}{{\log }_{a}}b$.
$\begin{align}
  & {{\log }_{{{x}^{2}}}}\left( {{2}^{4}} \right)+{{\log }_{2x}}\left( {{2}^{6}} \right)=3 \\
 & \Rightarrow \dfrac{4}{2}{{\log }_{x}}2+6{{\log }_{2x}}2=3 \\
 & \Rightarrow 2{{\log }_{x}}2+6{{\log }_{2x}}2=3 \\
\end{align}$
We have the relation of logarithm ${{\log }_{a}}b=\dfrac{1}{{{\log }_{b}}a}$.applying this theorem we get
$\begin{align}
  & 2{{\log }_{x}}2+6{{\log }_{2x}}2=3 \\
 & \Rightarrow \dfrac{2}{{{\log }_{2}}x}+\dfrac{6}{{{\log }_{2}}2x}=3 \\
\end{align}$
Now that we have changed the base value of the logarithm, we try to break the denominator.
We apply the theorem of ${{\log }_{a}}\left( bc \right)={{\log }_{a}}b+{{\log }_{a}}c$ and ${{\log }_{a}}a=1$.
\[\begin{align}
  & \dfrac{2}{{{\log }_{2}}x}+\dfrac{6}{{{\log }_{2}}2x}=3 \\
 & \Rightarrow \dfrac{2}{{{\log }_{2}}x}+\dfrac{6}{{{\log }_{2}}x+{{\log }_{2}}2}=3 \\
 & \Rightarrow \dfrac{2}{{{\log }_{2}}x}+\dfrac{6}{{{\log }_{2}}x+1}=3 \\
\end{align}\]
Now we assume a variable m for ${{\log }_{2}}x$ where ${{\log }_{2}}x=m$.
\[\begin{align}
  & \dfrac{2}{{{\log }_{2}}x}+\dfrac{6}{{{\log }_{2}}x+1}=3 \\
 & \Rightarrow \dfrac{2}{m}+\dfrac{6}{m+1}=3 \\
\end{align}\]
We have a quadratic equation of m. we first form the equation.
\[\begin{align}
  & \dfrac{2}{m}+\dfrac{6}{m+1}=3 \\
 & \Rightarrow \dfrac{2\left( m+1 \right)+6m}{m\left( m+1 \right)}=3 \\
 & \Rightarrow 2m+2+6m=3m\left( m+1 \right) \\
 & \Rightarrow 3{{m}^{2}}-5m-2=0 \\
\end{align}\]
Now we solve it using factorisation method
\[\begin{align}
  & 3{{m}^{2}}-5m-2=0 \\
 & \Rightarrow 3{{m}^{2}}-6m+m-2=0 \\
 & \Rightarrow \left( m-2 \right)\left( 3m+1 \right)=0 \\
\end{align}\]
The solution of the factors is $m=\dfrac{-1}{3},2$.
Now we need to replace back ${{\log }_{2}}x=m$. This gives us ${{\log }_{2}}x=\dfrac{-1}{3},2$.
We use the logarithmic formula ${{\log }_{a}}b=c\Rightarrow {{a}^{c}}=b$.
So, when ${{\log }_{2}}x=\dfrac{-1}{3}$, we get $x={{2}^{\dfrac{-1}{3}}}$. Now, the cube root of 2 contains two imaginary roots and one real root.
When ${{\log }_{2}}x=2$, we get $x={{2}^{2}}=4$. This has only one real root.
So, in total we got 2 real solutions and 2 imaginary solutions.

So, the correct answer is “Option C and D”.

Note: We need to remember that the number of roots is equal to the indices value. When we were trying to find the value of $x={{2}^{\dfrac{-1}{3}}}$, we were finding cube roots. It has 3 in the indices which gave us 3 roots of the solution.