A) \[x\in [0,81]\]
B) \[x\in \left( 0,{{3}^{72}} \right)\]
C) \[x\in [0,64)\]
D) \[-\,64\le x\le +\,64\]
Correct Answer: C
Solution :
(c): The given in equation is valid only when \[x\ge 0\]................(1) The given in equation can be written in the form. \[{{3}^{72-x-\sqrt{x}}}>1\] \[\Rightarrow 72-x-\sqrt{x}>0\,\,\,\,\left( \therefore 3{}^\circ =1 \right)\] \[\Rightarrow x+\sqrt{x}-72<0\] \[\Rightarrow \]\[\left( \sqrt{x}+9 \right)\left( \sqrt{x}-8 \right)<0\] But, \[\sqrt{x}+9>0\] for all \[x\ge 0\] So, \[\sqrt{x}-8<0\Rightarrow \sqrt{x}<8\] \[\therefore 0\le x<64\] [from (1)]
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