Question

Find out the solution of the equation ${{x}^{\left( \dfrac{3}{4} \right){{\left( \log x \right)}^{2}}+\log x-\dfrac{5}{4}}}=\sqrt{2}$. Indicate all the alternatives.
A. at least one real solution
B. exactly three real solutions
C. exactly one irrational solution.
D. complex roots

Answer 1

Hint:To simplify the solution we take logarithm both sides. We get the equation of log x. we get a cubic equation. We solve it using the factorization method. We get values of log x. We get values of x using logarithmic operations to complete the solution.

Complete step-by-step solution:
We first try to minimize the complexity of the equation by solving it.
We apply methods of logarithm in ${{x}^{\left( \dfrac{3}{4} \right){{\left( \log x \right)}^{2}}+\log x-\dfrac{5}{4}}}=\sqrt{2}$.
We take logarithm both sides to get $\log \left( {{x}^{\left( \dfrac{3}{4} \right){{\left( \log x \right)}^{2}}+\log x-\dfrac{5}{4}}} \right)=\log \left( \sqrt{2} \right)$.
We know that $\log \left( {{a}^{b}} \right)=b\log a$ and ${{\log }_{a}}a=1$. So, ${{\log }_{2}}2=1$.
The equation has all logarithm with base 2.
$\begin{align}
  & \left( \left( \dfrac{3}{4} \right){{\left( \log x \right)}^{2}}+\log x-\dfrac{5}{4} \right)\times \log \left( x \right)=\dfrac{1}{2}\log 2 \\
 & \Rightarrow 3{{\left( \log x \right)}^{3}}+4{{\left( \log x \right)}^{2}}-5\log x=2\times 1=2 \\
 & \Rightarrow 3{{\left( \log x \right)}^{3}}+4{{\left( \log x \right)}^{2}}-5\log x-2=0 \\
\end{align}$
We get a cubic equation of log x. We solve it by using vanishing method.
We take value of log x as $\log x=1$.
So, $\log x-1=0$ becomes a root of the equation.
$\begin{align}
  & 3{{\left( \log x \right)}^{3}}+4{{\left( \log x \right)}^{2}}-5\log x-2=0 \\
 & \Rightarrow \left( \log x-1 \right)\left[ 3{{\left( \log x \right)}^{2}}+7\log x+2 \right]=0 \\
\end{align}$
From the solution we get two values of log x.
So, $\log x=1$ and the other roots of log x are
$\begin{align}
  & 3{{\left( \log x \right)}^{2}}+7\log x+2 \\
 & \Rightarrow \log x=\dfrac{-7\pm \sqrt{{{7}^{2}}-4\times 3\times 2}}{2\times 3}=\dfrac{-7\pm 5}{6} \\
 & \Rightarrow \log x=-2,-\dfrac{1}{3} \\
\end{align}$
So, we get 3 possible values od log x as $\log x=-2,-\dfrac{1}{3},1$, all of which are possible in real domain.
Values of x will be $x={{2}^{-2}},{{2}^{-\dfrac{1}{3}}},{{2}^{1}}=\dfrac{1}{4},\dfrac{1}{\sqrt[3]{2}},2$ which we got from the formula of ${{\log }_{a}}b=x\Rightarrow {{a}^{x}}=b$.
So, there are exactly 3 real solutions of x. the correct option is (B).

Note: We can also use graph to solve the problem. But that will become more advanced level solving. We use the power of x in the equation and put it in the graph. We get the maximum and minimum values to find out its range. Then we equate with the right-hand side value to find the intersecting points.