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

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(c) ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{3 \over 2}} \right)^{2 - 2x}}$ ==> ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{2 \o

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(c) ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{3 \over 2}} \right)^{2 - 2x}}$ ==> ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{2 \over 2}} \right)^{2 - 2x}}$.

Clearly $x + 2 = 2x - 2$ ==> $x = 4$

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

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