A) 0
B) 3
C) both (a) and (b)
D) 0 and 6
Correct Answer: A
Solution :
\[{{\log }_{2}}(9-{{2}^{x}})={{10}^{{{\log }_{10}}(3-x)}}\] \[\Rightarrow \] \[{{\log }_{2}}(9-{{2}^{x}})=(3-x)\] \[[\because \,\,{{a}^{{{\log }_{a}}b}}=b]\] \[\Rightarrow \] \[{{2}^{3-x}}=9-{{2}^{x}}\] \[\Rightarrow \] \[\frac{{{2}^{3}}}{{{2}^{x}}}=9-{{2}^{x}}\] \[\Rightarrow \] \[8={{2}^{x}}\times (9-{{2}^{x}})\] \[\Rightarrow \] \[{{2}^{2x}}-{{2}^{x}}\times 9+8=0\] Let\[{{2}^{x}}=y\], then; \[{{y}^{2}}-9y+8=0\] \[\Rightarrow \] \[(y-8)(y-1)=0\] \[\Rightarrow \] \[y=8\] or \[y=1\] \[\Rightarrow \] \[{{2}^{x}}={{2}^{3}}\] or \[{{2}^{x}}={{2}^{0}}\] \[\Rightarrow \] \[x=3\] or \[x=0\]. But\[x=3\]does not satisfy the given equation, since\[\log 0\]is not defined.
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