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

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(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

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$4$

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(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 $x =$

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