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Basic of Logarithms
medium
Solution of the equation ${4.9^{x - 1}} = 3\sqrt {({2^{2x + 1}})} $ has the solution
A
$3$
B
$2$
C
$1.5$
D
$2/3$
Solution
(c) $4.\,{9^{x – 1}} = 3\,.\,\sqrt {({2^{2x + 1}})} $$ \Rightarrow $${3^{2x – 2 – 1}} = {2^{{{2x + 1} \over 2} – 2}}$
$ \Rightarrow $ ${3^{2x – 3}} = {2^{{{2x – 3} \over 2}}}$ $ \Rightarrow $ ${2^{{{2x – 3} \over 2}}} = {\left( {{3^{{{2x – 3} \over 2}}}} \right)^2}$
$ \Rightarrow $$2x – 3 = 0$,
$\therefore x = {3 \over 2}$.
Standard 11
Mathematics
