| If \[{{x}^{x\sqrt{x}}}={{(x\sqrt{x})}^{x}},\] then x is equal to [SSC (CGL) 2005] |
A) \[\frac{4}{9}\]
B) \[\frac{2}{3}\]
C) \[\frac{9}{4}\]
D) \[\frac{3}{2}\]
Correct Answer: C
Solution :
| \[{{x}^{x\sqrt{x}}}={{(x\sqrt{x})}^{x}}\]\[\Rightarrow \]\[{{x}^{x.{{x}^{1/2}}}}={{(x.{{x}^{1/2}})}^{x}}\] |
| \[\Rightarrow \]\[{{x}^{x\,\,\left( 1+\frac{1}{2} \right)}}={{({{x}^{1\,\,+\,\,1/2}})}^{x}}\]\[\Rightarrow \]\[{{x}^{{{x}^{3/2}}}}={{({{x}^{3/2}})}^{x}}={{x}^{3x/2}}\] |
| \[\Rightarrow \]\[{{x}^{{{x}^{3/2}}}}={{x}^{3x/2}}\] |
| Base is same. |
| \[\therefore \] \[{{x}^{3/2}}=\frac{3x}{2}\] |
| \[\Rightarrow \]\[{{x}^{3/2}}-\frac{3x}{2}=0\]\[\Rightarrow \]\[x\left( {{x}^{\frac{1}{2}}}-\frac{3}{2} \right)=0\]\[\Rightarrow \]\[x=0\] |
| or \[{{x}^{1/2}}=\frac{3}{2}\]\[\Rightarrow \]\[x={{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{4}\] |
| \[x=0\] given indeterminate value. |
| \[\therefore \]\[x=\frac{9}{4}\] |
You need to login to perform this action.
You will be redirected in
3 sec
