Question

If ${\log }_2(2^{x-1}+6)+{\log }_2(4^{x-1})=5,$ then $x=$

• A. 4
• B. 1
• C. 3
• D. 2

Answer 1

Answer: (D)

We have, ${\log }_2(2^{x-1}+6)+{\log }_2(4^{x-1})=5{\log }_2[(2^{x-1}+6)4^{x-1}]=5$ .
$\Rightarrow (2^{x-1}+6)4^{x-1}=2s\Rightarrow (2^{x-1}+6)(2^{x-1}{)}^2=32$ $\Rightarrow (t+6)t^2=32(put2x-l=t)$ $\Rightarrow t^3+6t^2-32=0\Rightarrow t=2\Rightarrow 2^{x-1}=2$ $\Rightarrow x-1=1\Rightarrow x=2$