A) 2
B) 5
C) 3
D) 1
Correct Answer: B
Solution :
We have \[{{2}^{x-3}}{{.3}^{2x-8}}=36\] \[\Rightarrow \]\[{{2}^{x}}{{2}^{-3}}{{.3}^{2x}}{{3}^{-8}}=36\] \[\Rightarrow \]\[{{2}^{x}}.\frac{1}{{{2}^{3}}}{{.3}^{x}}{{.3}^{x}}.\frac{1}{{{3}^{8}}}=36\] \[\Rightarrow \]\[{{(2\times 3\times 3)}^{x}}=36\times 8\times 6561\] \[\Rightarrow \]\[{{(18)}^{x}}={{(18)}^{5}}\] On comparing. We get \[\,x=5\]
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