If {x^{xsqrt[3]{x}}} = {(x cdot sqrt[3]{x})^x | vedclass

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

Question

(D) Given the equation: ${x^{x \cdot x^{1/3}}} = {(x \cdot x^{1/3})^x}$.Simplify the exponents: ${x^{x^{4/3}}} = {(x^{4/3})^x}$.This simplifies to: ${

Vedclass · Accepted Answer

$64/27$

Vedclass · Answer

(D) Given the equation: ${x^{x \cdot x^{1/3}}} = {(x \cdot x^{1/3})^x}$.Simplify the exponents: ${x^{x^{4/3}}} = {(x^{4/3})^x}$.This simplifies to: ${x^{x^{4/3}}} = {x^{(4/3)x}}$.For the bases to be equal,the exponents must be equal (assuming $x > 0$ and $x 
eq 1$):${x^{4/3}} = \frac{4}{3}x$.Divide by $x$ (since $x 
eq 0$):${x^{1/3}} = \frac{4}{3}$.Cube both sides:$x = (\frac{4}{3})^3 = \frac{64}{27}$.

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

Similar Questions

The solution of the equation $4 \cdot 9^{x - 1} = 3 \cdot \sqrt{2^{2x + 1}}$ is

The value of $\frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}}$ is

$\frac{\sqrt{5/2} + \sqrt{7 - 3\sqrt{5}}}{\sqrt{7/2} + \sqrt{16 - 5\sqrt{7}}} = $

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

If ${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{3}{2} \right)^{2 - 2x}}$,then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module