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Basic of Logarithms
easy
જો ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{3 \over 2}} \right)^{2 - 2x}},$ તો $x =$
A
$1$
B
$3$
C
$4$
D
$0$
Solution
(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 $x + 2 = 2x – 2$ ==> $x = 4$
Standard 11
Mathematics
