The solution to the equation (x)^{xsqrt{x}} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation: $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$We can rewrite the right side as: $(x \cdot x^{1/2})^x = (x^{3/2})^x = x^{3x/2}$So,the equati

Vedclass · Accepted Answer

$9/4$

Vedclass · Answer

(A) Given the equation: $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$We can rewrite the right side as: $(x \cdot x^{1/2})^x = (x^{3/2})^x = x^{3x/2}$So,the equation becomes: $x^{x\sqrt{x}} = x^{3x/2}$This implies $x^{x^{3/2}} = x^{3x/2}$Equating the exponents: $x^{3/2} = \frac{3x}{2}$Dividing by $x$ (assuming $x 
eq 0$): $x^{1/2} = \frac{3}{2}$Squaring both sides: $x = (\frac{3}{2})^2 = \frac{9}{4}$Thus,the solution is $x = \frac{9}{4}$.

The solution to the equation $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$ is:

Similar Questions

$({x^5})^{1/3} \times (16{x^3})^{2/3} \times \left( \frac{1}{4}{x^{4/9}} \right)^{-3/2} = ?$

If $2^x = 4^y = 8^z$ and $xyz = 288$,then $\frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = $

Which of the following statements are true and which are false? In each case,give a valid reason for your answer.
$t: \sqrt{11}$ is a rational number.

If ${x^{x\sqrt[3]{x}}} = {(x \cdot \sqrt[3]{x})^x}$,then $x =$

If $x = 3 - \sqrt{5}$,then find the value of $\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module