Question

solve for x.

${ \tt \ \: |2 - log_{ \frac{1}{2} }(x) | = 3 + log_{2}(x - 1)}$

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Answer 1

Answer:

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Answer 2

If $f(x)=|x|,$

$f(x) = \begin{cases} - x \: \: \: \: x < 0 \\ x \: \: \: \: \: \: x \geqslant 0 \end{cases}$

Given that,

$|2 - \log_{ \frac{1}{2} }(x) | = 3 + \log_{2}(x - 1)$

Since $\log(b)$ is defined only for $b>0,$

From LHS,

$x>0$

and from RHS,

$(x-1)>0$

$\longrightarrow x>1$

Taking the intersection,

$\longrightarrow \underline{x \in (1,\infty)}\quad \quad\dots(1)$

Case 1:

$2 - \log_{ \frac{1}{2} }(x) < 0$

$\longrightarrow 2 + \log_{ 2 }(x) < 0$

$\longrightarrow \log_{ 2 }(x) < - 2$

$\longrightarrow x < {2}^{ - 2}$

$\longrightarrow x < \dfrac{1}{4} \quad \quad\dots(2)$

Then,

$- \bigg \{2 - \log_{ \frac{1}{2} }(x) \bigg \} = 3 + \log_{2}(x - 1)$

$\longrightarrow - 2 + \log_{ \frac{1}{2} }(x) = 3 + \log_{2}(x - 1)$

$\longrightarrow \log_{2}(x - 1) - \log_{ \frac{1}{2} }(x) = - 5$

$\longrightarrow \log_{2}(x - 1) + \log_{2}(x) = - 5$

$\longrightarrow \log_{2}x(x - 1) = - 5$

$\longrightarrow {2}^{ - 5} = x(x - 1)$

$\longrightarrow \dfrac{1}{32} = {x}^{2} - x$

$\longrightarrow 32 {x}^{2} - 32x - 1 = 0$

On solving the quadratic equation, we get,

$\longrightarrow x_{1} = \dfrac{4 + 3 \sqrt{2} }{8} \approx1.0303\quad \quad\dots(3)$

$\longrightarrow x_{2} = \dfrac{4 - 3 \sqrt{2} }{8} \approx - 0.0303\quad \quad\dots(4)$

Taking intersection of (2), (3) and (4), the only solution we get for case 1 is,

$x_{2} = \dfrac{4 - 3 \sqrt{2} }{8} \approx - 0.0303$

Case 2:

$2 - \log_{ \frac{1}{2} }(x) \geqslant 0$

$\longrightarrow 2 + \log_{ 2 }(x) \geqslant 0$

$\longrightarrow \log_{ 2 }(x) \geqslant - 2$

$\longrightarrow x \geqslant {2}^{ - 2}$

$\longrightarrow x \geqslant \dfrac{1}{4} \quad \quad\dots(5)$

Then,

$2 - \log_{ \frac{1}{2} }(x) = 3 + \log_{2}(x - 1)$

$\longrightarrow 2 + \log_{ 2 }(x) = 3 + \log_{2}(x - 1)$

$\longrightarrow \log_{ 2 }(x) - \log_{2}(x - 1) = 1$

$\longrightarrow \log_{ 2 } \bigg( \dfrac{x}{x - 1} \bigg) = 1$

$\longrightarrow 2 = \dfrac{x}{x - 1}$

$\longrightarrow x_{3} = 2\quad \quad\dots(6)$

Taking the intersection of (5) and (6), the solution for case 2 is,

$\longrightarrow x_{3} = 2$

But considering (1), we can eliminate $x_2.$

Hence the only solution of

$|2 - \log_{ \frac{1}{2} }(x) | = 3 + \log_{2}(x - 1)$

is,

$\longrightarrow \underline{ \underline{x = 2}}$