Q79 / 180 · COMEDK 2018 · Mathematics

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

  • A
    4
  • B
    1
  • C
    3
  • Correct answer
    D
    2

Solution

We have, $$ \begin{aligned} &\log _{2}\left(2^{x-1}+6\right)+\log _{2}\left(4^{x-1}\right)=5 \\ &\left.\Rightarrow 2^{x-1}+6\right)\left(2^{2 x-2}\right)=2^{5} \end{aligned} $$ Let $$ y=2^{x-1} $$ $$ \Rightarrow \quad \cdot(y+6) y^{2}=2^{5} $$ $$ \begin{aligned} &\Rightarrow \quad y^{3}+6 y^{2}-32=0 \\ &\Rightarrow \quad(y-2)\left(y^{2}+8 y+16\right)=0 \end{aligned} $$ So, $$ \begin{aligned} y &=2 \\ 2^{x-1} &=2^{1} \end{aligned} $$ On comparing both sides, we get $$ \begin{array}{r} x-1=1 \\ x=2 \end{array} $$

Hence, the correct answer is option: 'D'.

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