Question

If $ {{x}^{x\sqrt{x}}}={{\left( x\sqrt{x} \right)}^{x}} $ , find the value of x.
A. $ \dfrac{3}{2} $
B. $ \dfrac{2}{9} $
C. $ \dfrac{9}{4} $
D. $ \dfrac{4}{9} $

Answer 1

Hint: In the given question, we will use the logarithm function and its properties to find the required value of x. We will take log to the base x to both sides of the given equation and then we will use the following property, $ {{\log }_{x}}{{x}^{m}}=m{{\log }_{x}}x=m\left( 1 \right) $ .

Complete step-by-step answer:
We have been given $ {{x}^{x\sqrt{x}}}={{\left( x\sqrt{x} \right)}^{x}} $ and have been asked to find the value of x. To solve this question, we will first take log on both sides of the given equation to the base x. So, we will get,
 $ {{\log }_{x}}{{x}^{x\sqrt{x}}}={{\log }_{x}}{{\left( x\sqrt{x} \right)}^{x}} $
Now, we know that $ {{\log }_{x}}{{x}^{m}}=m{{\log }_{x}}x=m\left( 1 \right) $ . We will apply this in the above equation. So, we get the above equation as,
 $ x\sqrt{x}{{\log }_{x}}x=x{{\log }_{x}}x\sqrt{x} $
We know that $ \sqrt{a}={{a}^{\dfrac{1}{2}}} $ . So, we will apply that in the above equation. So, we will get,
 $ x\sqrt{x}=x{{\log }_{x}}\left( x\times {{x}^{\dfrac{1}{2}}} \right) $
We will now divide the equation by x, so we get,
 $ \sqrt{x}={{\log }_{x}}\left( {{x}^{\left( 1+\dfrac{1}{2} \right)}} \right) $
We know that $ {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} $ . So, we will apply this in the above equation and get,
 $ \sqrt{x}={{\log }_{x}}{{x}^{\dfrac{3}{2}}} $
We will again use the property of $ {{\log }_{x}}{{x}^{m}}=m{{\log }_{x}}x=m\left( 1 \right) $ . So, we will get as,
 $ \sqrt{x}=\dfrac{3}{2} $
We will now square both sides of the equation. So, we will get the equation as,
 $ \begin{align}
  & {{\left( \sqrt{x} \right)}^{2}}={{\left( \dfrac{3}{2} \right)}^{2}} \\
 & \Rightarrow x=\dfrac{9}{4} \\
\end{align} $
Hence, we get the value of x as $ \dfrac{9}{4} $ .
Therefore, the correct option is option C.

Note: We should be careful while using the logarithmic functions and its properties as there are chances of minor mistakes while solving the equation. We have to remember that we can solve this type of question by taking log and using the property of logarithmic function. If the student does not know that $ {{\log }_{x}}x=1 $ , then it might be difficult to simplify and solve this question. So, students know that $ \log 10=1 $ , they can recollect that $ {{\log }_{10}}10=1 $ . So, generalising this in terms of x, we get that $ {{\log }_{x}}x=1 $ .