A) 24
B) 26
C) 28
D) 30
Correct Answer: B
Solution :
| We have, \[{{\log }_{2x}}1944={{\log }_{x}}486\sqrt{2}\]\[\Rightarrow \frac{{{\log }_{2}}1944}{{{\log }_{2}}2+{{\log }_{2}}x}=\frac{{{\log }_{2}}486\sqrt{2}}{{{\log }_{2}}x}\]\[\Rightarrow {{\log }_{2}}x.{{\log }_{2}}1944={{\log }_{2}}486\sqrt{2}\left( 1+{{\log }_{2}}x \right)\] |
| \[\Rightarrow {{\log }_{2}}x({{\log }_{2}}1944-{{\log }_{2}}486\sqrt{2})\]\[={{\log }_{2}}486\sqrt{2}\]\[\Rightarrow {{\log }_{2}}x\left( {{\log }_{2}}\frac{1944}{486\sqrt{2}} \right)={{\log }_{2}}486\sqrt{2}\]\[\Rightarrow \,\,\,{{\log }_{2}}x{{\log }_{2}}2\sqrt{2}={{\log }_{2}}486\sqrt{2}\]\[\Rightarrow \frac{3}{2}{{\log }_{2}}x={{\log }_{2}}486\sqrt{2}\]\[\Rightarrow {{\log }_{2}}{{x}^{3/2}}={{\log }_{2}}486\sqrt{2}\]\[\Rightarrow {{x}^{3/2}}=486\sqrt{2}\]\[\Rightarrow {{x}^{6}}={{2}^{6}}\times {{3}^{20}}\] |
| \[\therefore \,\,a=6,b=20\] |
| \[a+b=26\] |
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