A) \[\frac{4}{3}\]
B) \[\frac{3}{2}\]
C) \[\frac{2}{1}\]
D) \[\frac{5}{3}\]
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 \[{{2}^{2x}}\left( 1+\frac{1}{2} \right)={{3}^{x-\frac{1}{2}}}(1+3)\] 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 \[x\log \left( \frac{4}{3} \right)=\log 8-\log 3\sqrt{3}\] Þ \[{{\left( \frac{4}{3} \right)}^{x}}=\frac{8}{3\sqrt{3}}\] Þ \[{{\left( \frac{4}{3} \right)}^{x}}={{\left( \frac{4}{3} \right)}^{3/2}}\] \[\therefore \,\,x=\frac{3}{2}\] Trick: Cheak the equation with options then only option satisfies the equation.
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