Question

If \[{{x}_{1}}\] and \[{{x}_{2}}\] are roots of equation \[{{e}^{\dfrac{3}{2}}}\cdot {{x}^{2\ln x}}={{x}^{4}}\]. Then the product of the roots of the equation is?

Answer 1

Hint: In order to find the product of roots of the given equation, we will be applying logarithm on both sides of the equation. Then we will be solving it by applying and following the rules of logarithm and we will be obtaining the roots of the equation. After obtaining the roots, we will be finding the product of the roots and that would be our required answer.

Complete step by step answer:
Now let us learn about the logarithms. It is just the exponentiation of another number. Generally, the logarithms are written to the base of 10. There are two types of logarithms. They are: Natural logarithms and Common logarithms. Logarithms follow the product rule, division rule, power rule, change of base rule, derivative of log , integral of log and etc. the logarithm of 0 is undefined.
Now let us find the roots of the given equation i.e. \[{{e}^{\dfrac{3}{2}}}\cdot {{x}^{2\ln x}}={{x}^{4}}\]
Firstly, we will be applying \[{{\log }_{e}}\] on both sides. We get,
\[\log \left( {{e}^{\dfrac{3}{2}}}\cdot {{x}^{2\ln x}} \right)=\log {{x}^{4}}\]
\[\Rightarrow \dfrac{3}{2}+2\times \log x\times \log x=4\log x\] \[\because \left[ \log {{m}^{n}}=n\log m \right]\], \[\log_{e}x\]=lnx and \[\left[ \log {{e}^{x}}=x \right]\]
Upon further solving, we get
\[\Rightarrow 2{{\left( \log x \right)}^{2}}-4\log x+\dfrac{3}{2}=0\]
Now for easy calculation, let us multiply with 2 the entire equation. We get
\[\Rightarrow 4{{\left( \log x \right)}^{2}}-8\log x+3=0\]
\[\begin{align}
  & \Rightarrow \left( \left( \log x \right)-\dfrac{8\pm \sqrt{64-48}}{8} \right)=0 \\
 & \Rightarrow \left[ \log x-\dfrac{8\pm 4}{8} \right]=0 \\
 & \Rightarrow \left( \log x-\dfrac{3}{2} \right)\left( \log x-\dfrac{1}{2} \right)=0 \\
 & \Rightarrow x={{e}^{\dfrac{3}{2}}},{{e}^{\dfrac{1}{2}}} \\
\end{align}\]
\[\therefore \] The roots of the equation \[{{e}^{\dfrac{3}{2}}}\cdot {{x}^{2\ln x}}={{x}^{4}}\] is \[{{e}^{\dfrac{3}{2}}},{{e}^{\dfrac{1}{2}}}\].
The product of the roots are
\[\Rightarrow a={{e}^{\dfrac{3}{2}}}\times {{e}^{\dfrac{1}{2}}}={{e}^{\dfrac{3+1}{2}}}={{e}^{2}}\]

Note: While solving the logarithmic problems, we must be aware of all the logarithmic rules and formulas. We must try and solve each part by applying the rules as we did in the above problem. Logarithms can be applied widely in science and technology.