A) \[\frac{1}{6}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{2}\]
D) \[6\]
Correct Answer: C
Solution :
| We have, \[{{\left( 2x \right)}^{ln\,2}}={{\left( 3y \right)}^{ln\,3}}\]\[\Rightarrow \,\,\,\,ln2\left( ln\,\,2+ln\,x \right)=ln\,3\left( ln\,3+ln\,y \right)\] | |
| \[\Rightarrow \,{{\left( ln\,\,2 \right)}^{2}}+ln\,2.\,ln\,x={{\left( ln\,3 \right)}^{2}}+ln\,3.lny\] | ?? (i) \[\Rightarrow \,\,\,\,\,\,{{3}^{ln\,x}}={{2}^{ln\,y}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,ln\,\,x\,In\,\,3=ln\,\,y\,ln\,\,2\] | ?? (ii) |
| From Eqs. (i) and (ii) , we get | |
| \[{{\left( ln\,2 \right)}^{2}}+In\,2.\,lnx\]\[={{\left( ln\,\,3 \right)}^{2}}+ln\,3\left( \frac{ln\,x.\,ln\,3}{ln\,2} \right)\]\[\begin{align} & \Rightarrow \,\,{{\left( ln\,2 \right)}^{3}}+{{\left( ln\,2 \right)}^{2}}ln\,x={{\left( ln\,3 \right)}^{2}}ln\,2 \\ & +{{\left( ln\,3 \right)}^{2}}ln\,x \\ \end{align}\] | |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,(\ln \,x)\,\,[{{(\ln \,3)}^{2}}-{{(\ln \,2)}^{2}}]\]\[=\ln \,2\,[{{(\ln \,2)}^{2}}-{{(\ln \,3)}^{2}}]\]\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,ln\,\,x=-ln\,2\] | |
| \[\therefore \,\,\,\,\,\,\,x=\frac{1}{2}\] | |
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