Solve : (log)2x−2x−1>0

Solution: log2(x−2x−1)>0⇒x−2x−1>20⇒x−2x−1>1⇒x−2x−1−1>0⇒x−2(x−1)−(x−2)>0⇒x−21>0It will be possible, when x>2.So, x∈(2,∞) is the solution for this equation.

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Solve : {({})}_{{2}}{{{x}-{1}}}{{{x}-{2}}}>{0}

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Solve : $(lo g)_{2} \frac{x - 1}{x - 2} > 0$

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Text solutionVerified

Solution:

lo g_{2} (\frac{x - 1}{x - 2}) > 0

\Rightarrow \frac{x - 1}{x - 2} > 2^{0}

\Rightarrow \frac{x - 1}{x - 2} > 1

\Rightarrow \frac{x - 1}{x - 2} - 1 > 0

\Rightarrow \frac{( x - 1 ) - ( x - 2 )}{x - 2} > 0

\Rightarrow \frac{1}{x - 2} > 0

It will be possible, when

x > 2

.
So,

x \in (2, \infty)

is the solution for this equation.

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Question Text	Solve : $(lo g)_{2} \frac{x - 1}{x - 2} > 0$
Answer Type	Text solution:1
Upvotes	150

Solve : (log)2​x−2x−1​>0

Text solutionVerified

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Solve : $(lo g)_{2} \frac{x - 1}{x - 2} > 0$