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If log(1//10)(x^2+x) gtlog(1//2)(x^3-x) ...

If `log_(1//10)(x^2+x) gtlog_(1//2)(x^3-x) +log_(2)(x-1) `, then :

A

`x in (1,oo)`

B

`x in ((sqrt5-1)/2,1)`

C

`x in ((-sqrt5-1)/2,(sqrt5-1)/2)`

D

`x in (-oo,(-sqrt5-1)/2)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the inequality \( \log_{(1/10)}(x^2 + x) > \log_{(1/2)}(x^3 - x) + \log_{2}(x - 1) \), we will follow these steps: ### Step 1: Determine the conditions for the logarithmic expressions to be defined For the logarithmic expressions to be defined, the arguments of the logarithms must be positive. 1. **For \( \log_{(1/10)}(x^2 + x) \)**: \[ x^2 + x > 0 \] Factoring gives: \[ x(x + 1) > 0 \] The critical points are \( x = 0 \) and \( x = -1 \). Testing intervals: - For \( x < -1 \): both factors are negative, product is positive. - For \( -1 < x < 0 \): one factor is negative, product is negative. - For \( x > 0 \): both factors are positive, product is positive. Thus, \( x(x + 1) > 0 \) for \( x \in (-\infty, -1) \cup (0, \infty) \). 2. **For \( \log_{(1/2)}(x^3 - x) \)**: \[ x^3 - x > 0 \] Factoring gives: \[ x(x^2 - 1) > 0 \quad \Rightarrow \quad x(x - 1)(x + 1) > 0 \] The critical points are \( x = -1, 0, 1 \). Testing intervals: - For \( x < -1 \): all factors are negative, product is negative. - For \( -1 < x < 0 \): one factor is negative, product is positive. - For \( 0 < x < 1 \): one factor is negative, product is negative. - For \( x > 1 \): all factors are positive, product is positive. Thus, \( x(x - 1)(x + 1) > 0 \) for \( x \in (-1, 0) \cup (1, \infty) \). 3. **For \( \log_{2}(x - 1) \)**: \[ x - 1 > 0 \quad \Rightarrow \quad x > 1 \] ### Step 2: Combine the intervals Now we combine the intervals where all logarithmic expressions are defined: - From \( \log_{(1/10)}(x^2 + x) \): \( (-\infty, -1) \cup (0, \infty) \) - From \( \log_{(1/2)}(x^3 - x) \): \( (-1, 0) \cup (1, \infty) \) - From \( \log_{2}(x - 1) \): \( (1, \infty) \) The common interval that satisfies all conditions is: \[ (1, \infty) \] ### Step 3: Solve the inequality Now we need to solve the original inequality. Since we have already established that \( x \) must be in \( (1, \infty) \), we can proceed to analyze the inequality within this interval. ### Final Result The solution to the inequality \( \log_{(1/10)}(x^2 + x) > \log_{(1/2)}(x^3 - x) + \log_{2}(x - 1) \) is: \[ x \in (1, \infty) \]
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