Search any question & find its solution
Question:
Answered & Verified by Expert
If $\log _{2}^{6}+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)$, then the values of $x$ are
Options:
Solution:
1821 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{4},\frac{1}{2}$
We have,
$\log _{2}^{6}+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)$
$\Rightarrow \quad 1+\log _{2}^{3}+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)$
$\Rightarrow \quad \frac{2^{\frac{1}{x}}+8}{3}=2^{1+\frac{1}{2 x}}$
$\Rightarrow 2^{\frac{1}{x}}+8=3 \cdot 2 \cdot 2^{\frac{1}{2 x}}$
$\Rightarrow 2^{\frac{1}{x}}+8=6 \cdot 2^{\frac{1}{2 x}}$
Let $y=2^{\frac{1}{2 x}}$
$\Rightarrow \quad y^{2}+8=6 y$
$\Rightarrow \quad y^{2}-6 y+8=0$
$\Rightarrow \quad(y-4)(y-2)=0$
$y=4,2$
$2^{\frac{1}{2 x}}=4$ and $2^{\frac{1}{2 x}}=2$
$\frac{1}{2 x}=2$ and $\frac{1}{2 x}=1$
$x=\frac{1}{4}, \frac{1}{2}$
$\log _{2}^{6}+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)$
$\Rightarrow \quad 1+\log _{2}^{3}+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)$
$\Rightarrow \quad \frac{2^{\frac{1}{x}}+8}{3}=2^{1+\frac{1}{2 x}}$
$\Rightarrow 2^{\frac{1}{x}}+8=3 \cdot 2 \cdot 2^{\frac{1}{2 x}}$
$\Rightarrow 2^{\frac{1}{x}}+8=6 \cdot 2^{\frac{1}{2 x}}$
Let $y=2^{\frac{1}{2 x}}$
$\Rightarrow \quad y^{2}+8=6 y$
$\Rightarrow \quad y^{2}-6 y+8=0$
$\Rightarrow \quad(y-4)(y-2)=0$
$y=4,2$
$2^{\frac{1}{2 x}}=4$ and $2^{\frac{1}{2 x}}=2$
$\frac{1}{2 x}=2$ and $\frac{1}{2 x}=1$
$x=\frac{1}{4}, \frac{1}{2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.