A B C D
A) 1 2 3 4
B) 3 4 2 1
C) 3 1 4 2
D) 1 2 4 3
Correct Answer: C
Solution :
(A) \[{{\left( \sqrt{6} \right)}^{x-3}}=1\] \[\Rightarrow \] \[{{6}^{\frac{x-3}{2}}}=1\,\,\,\Rightarrow {{6}^{\frac{x-3}{2}}}=6{}^\circ \] \[\Rightarrow \] \[\frac{x-3}{2}=0\,\,\,\Rightarrow x=3\] (B) \[{{2}^{2x+1}}={{4}^{2x-1}}\] \[\Rightarrow \] \[{{2}^{2x+1}}={{2}^{2(2x-1)}}\] \[\Rightarrow \] \[2x+1=4x-2\] \[\Rightarrow \] \[3=2x\Rightarrow x=\frac{3}{2}\] (C) \[{{9}^{x-1}}+18={{3}^{2x-1}}\] \[\Rightarrow \] \[{{9}^{x}}{{.9}^{-1}}+18={{({{3}^{2}})}^{x}}{{.3}^{-1}}\] \[\Rightarrow \] \[{{9}^{x}}.\frac{1}{9}+18={{9}^{x}}.\frac{1}{3}\] \[\Rightarrow \] \[{{9}^{x}}\left( \frac{1}{9}-\frac{1}{3} \right)=-18\] \[\Rightarrow \] \[x=2\] \[\Rightarrow \] \[1<x<4\] (D) \[{{\left( \frac{8}{3} \right)}^{2x+6}}={{\left( \frac{8}{3} \right)}^{x+2}}\] \[\Rightarrow \] \[2x+6=x+2\] \[\Rightarrow \] \[x=-4\]
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