If {left( frac{2}{3} right)^{x + 2}} = {left( | vedclass

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

Question

(C) Given equation: ${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{3}{2} \right)^{2 - 2x}}$We know that $\frac{3}{2} = {\left( \frac{2}{3} \rig

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given equation: ${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{3}{2} \right)^{2 - 2x}}$We know that $\frac{3}{2} = {\left( \frac{2}{3} \right)^{-1}}$.Substituting this into the equation: ${\left( \frac{2}{3} \right)^{x + 2}} = {\left( {\left( \frac{2}{3} \right)^{-1}} \right)^{2 - 2x}}$${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{2}{3} \right)^{-(2 - 2x)}}$${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{2}{3} \right)^{2x - 2}}$Since the bases are equal,we equate the exponents: $x + 2 = 2x - 2$$2 + 2 = 2x - x$$x = 4$

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

Similar Questions

The number of solutions of the equation $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$ is:

If $12^{4+2x^2} = (24\sqrt{3})^{3x^2-2}$,then $x$ is equal to

$\frac{12}{3 + \sqrt{5} - 2\sqrt{2}} = $

$\frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}$ is equal to

The rationalizing factor of $2\sqrt{3} - \sqrt{7}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module