Complex Numbers And Quadratic Equations question 745
Question: The value of x in the given equation $ 4^{x}-{3^{x\ -\ \frac{1}{2}}}={3^{x+\frac{1}{2}}}-{2^{2x-1}} $ is
Options:
A) $ \frac{4}{3} $
B) $ \frac{3}{2} $
C) $ \frac{2}{1} $
D) $ \frac{5}{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
Equation, $ 4^{x}-{3^{x-\frac{1}{2}}}={3^{x+\frac{1}{2}}}-{2^{2x-1}} $
Þ $ 2^{2x}+{2^{2x-1}}={3^{x+\frac{1}{2}}}+{3^{x-\frac{1}{2}}} $
Þ $ 2^{2x}( 1+\frac{1}{2} )={3^{x-\frac{1}{2}}}(1+3) $
Þ $ 2^{2x}.\frac{3}{2}={3^{x-\frac{1}{2}}}.4 $
Þ $ {2^{2x-3}}={3^{x-\frac{3}{2}}} $ Taking log both sides
Þ $ (2x-3)\log 2=(x-3/2)\log 3 $
Þ $ 2x\log 2-3\log 2=x\log 3-\frac{3}{2}\log 3 $
Þ $ x\log 4-x\log 3=3\log 2-\frac{3}{2}\log 3 $
Þ $ x\log ( \frac{4}{3} )=\log 8-\log 3\sqrt{3} $
Þ $ {{( \frac{4}{3} )}^{x}}=\frac{8}{3\sqrt{3}} $
Þ $ {{( \frac{4}{3} )}^{x}}={{( \frac{4}{3} )}^{3/2}} $
$ \therefore x=\frac{3}{2} $ Trick: Cheak the equation with options then only option satisfies the equation.
BETA
