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Sagot :
Sure, let's solve the inequality [tex]\( 1 + \log_2 x - 2 \log_x 2 > 0 \)[/tex] step-by-step.
### Step 1: Convert the logarithms to a common base
Start by expressing both logarithms in terms of natural logarithms. Recall that [tex]\( \log_b a = \frac{\ln a}{\ln b} \)[/tex].
So:
- [tex]\( \log_2 x = \frac{\ln x}{\ln 2} \)[/tex]
- [tex]\( \log_x 2 = \frac{\ln 2}{\ln x} \)[/tex]
### Step 2: Substitute these expressions into the inequality
Substituting into the original inequality, we get:
[tex]\[ 1 + \frac{\ln x}{\ln 2} - 2 \cdot \frac{\ln 2}{\ln x} > 0 \][/tex]
### Step 3: Simplify the inequality
Combine the terms to get a single inequality involving logarithms:
[tex]\[ 1 + \frac{\ln x}{\ln 2} - \frac{2 \ln 2}{\ln x} > 0 \][/tex]
### Step 4: Find a common denominator
To combine the terms more effectively, let's find a common denominator for the logarithms:
[tex]\[ \frac{\ln x \cdot \ln x + \ln 2 \cdot \ln x - 2 \ln 2^2}{\ln 2 \cdot \ln x} > 0 \][/tex]
Simplify the numerator:
[tex]\[ \frac{(\ln x)^2 + \ln 2 \cdot \ln x - 2 (\ln 2)^2}{\ln 2 \cdot \ln x} > 0 \][/tex]
### Step 5: Solve the simplified inequality
Now we set the numerator greater than 0 and solve:
[tex]\[ (\ln x)^2 + \ln 2 \cdot \ln x - 2 (\ln 2)^2 > 0 \][/tex]
This is a quadratic inequality in terms of [tex]\(\ln x\)[/tex]. Let [tex]\( t = \ln x \)[/tex], thus the inequality becomes:
[tex]\[ t^2 + \ln(2)t - 2(\ln(2))^2 > 0 \][/tex]
### Step 6: Solve the quadratic inequality
Solve [tex]\( t^2 + (\ln 2)t - 2(\ln 2)^2 = 0 \)[/tex] using the quadratic formula: [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = \ln(2) \)[/tex], and [tex]\( c = -2(\ln 2)^2 \)[/tex]:
[tex]\[ t = \frac{-\ln 2 \pm \sqrt{(\ln 2)^2 + 8(\ln 2)^2}}{2} \][/tex]
[tex]\[ t = \frac{-\ln 2 \pm \sqrt{9 (\ln 2)^2}}{2} \][/tex]
[tex]\[ t = \frac{-\ln 2 \pm 3 \ln 2}{2} \][/tex]
[tex]\[ t_1 = \frac{2 \ln 2}{2} = \ln 4 \][/tex]
[tex]\[ t_2 = \frac{-4 \ln 2}{2} = -2 \ln 2 \][/tex]
Since [tex]\( t = \ln x \)[/tex], we have:
[tex]\[ \ln x = \ln 4 \implies x = 4 \][/tex]
[tex]\[ \ln x = -2 \ln 2 \implies x = e^{-2 \ln 2} = \left(e^{\ln 2}\right)^{-2} = 2^{-2} = \frac{1}{4} \][/tex]
### Step 7: Investigate the intervals
Thus, the critical values are [tex]\( x = 4 \)[/tex] and [tex]\( x = \frac{1}{4} \)[/tex]. We need to test the intervals these points create:
1. [tex]\( 0 < x < \frac{1}{4} \)[/tex]
2. [tex]\( \frac{1}{4} < x < 1 \)[/tex]
3. [tex]\( 1 < x < 4 \)[/tex]
4. [tex]\( x > 4 \)[/tex]
On calculating the values in these intervals, especially considering that the solution must satisfy the inequality [tex]\( 1 + \log_2 x - 2 \log_x 2 > 0 \)[/tex], we find that the inequality holds for [tex]\( 2 < x \)[/tex] and [tex]\(\frac{1}{4} < x < 1 \)[/tex].
So, the solution to [tex]\( 1 + \log_2 x - 2 \log_x 2 > 0 \)[/tex] is:
[tex]\[ (2 < x) \cup \left(\frac{1}{4} < x < 1\right) \][/tex]
This means [tex]\( x \)[/tex] must be in the intervals:
[tex]\[ x \in \left(2, \infty\right) \cup \left(\frac{1}{4}, 1\right) \][/tex]
### Step 1: Convert the logarithms to a common base
Start by expressing both logarithms in terms of natural logarithms. Recall that [tex]\( \log_b a = \frac{\ln a}{\ln b} \)[/tex].
So:
- [tex]\( \log_2 x = \frac{\ln x}{\ln 2} \)[/tex]
- [tex]\( \log_x 2 = \frac{\ln 2}{\ln x} \)[/tex]
### Step 2: Substitute these expressions into the inequality
Substituting into the original inequality, we get:
[tex]\[ 1 + \frac{\ln x}{\ln 2} - 2 \cdot \frac{\ln 2}{\ln x} > 0 \][/tex]
### Step 3: Simplify the inequality
Combine the terms to get a single inequality involving logarithms:
[tex]\[ 1 + \frac{\ln x}{\ln 2} - \frac{2 \ln 2}{\ln x} > 0 \][/tex]
### Step 4: Find a common denominator
To combine the terms more effectively, let's find a common denominator for the logarithms:
[tex]\[ \frac{\ln x \cdot \ln x + \ln 2 \cdot \ln x - 2 \ln 2^2}{\ln 2 \cdot \ln x} > 0 \][/tex]
Simplify the numerator:
[tex]\[ \frac{(\ln x)^2 + \ln 2 \cdot \ln x - 2 (\ln 2)^2}{\ln 2 \cdot \ln x} > 0 \][/tex]
### Step 5: Solve the simplified inequality
Now we set the numerator greater than 0 and solve:
[tex]\[ (\ln x)^2 + \ln 2 \cdot \ln x - 2 (\ln 2)^2 > 0 \][/tex]
This is a quadratic inequality in terms of [tex]\(\ln x\)[/tex]. Let [tex]\( t = \ln x \)[/tex], thus the inequality becomes:
[tex]\[ t^2 + \ln(2)t - 2(\ln(2))^2 > 0 \][/tex]
### Step 6: Solve the quadratic inequality
Solve [tex]\( t^2 + (\ln 2)t - 2(\ln 2)^2 = 0 \)[/tex] using the quadratic formula: [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = \ln(2) \)[/tex], and [tex]\( c = -2(\ln 2)^2 \)[/tex]:
[tex]\[ t = \frac{-\ln 2 \pm \sqrt{(\ln 2)^2 + 8(\ln 2)^2}}{2} \][/tex]
[tex]\[ t = \frac{-\ln 2 \pm \sqrt{9 (\ln 2)^2}}{2} \][/tex]
[tex]\[ t = \frac{-\ln 2 \pm 3 \ln 2}{2} \][/tex]
[tex]\[ t_1 = \frac{2 \ln 2}{2} = \ln 4 \][/tex]
[tex]\[ t_2 = \frac{-4 \ln 2}{2} = -2 \ln 2 \][/tex]
Since [tex]\( t = \ln x \)[/tex], we have:
[tex]\[ \ln x = \ln 4 \implies x = 4 \][/tex]
[tex]\[ \ln x = -2 \ln 2 \implies x = e^{-2 \ln 2} = \left(e^{\ln 2}\right)^{-2} = 2^{-2} = \frac{1}{4} \][/tex]
### Step 7: Investigate the intervals
Thus, the critical values are [tex]\( x = 4 \)[/tex] and [tex]\( x = \frac{1}{4} \)[/tex]. We need to test the intervals these points create:
1. [tex]\( 0 < x < \frac{1}{4} \)[/tex]
2. [tex]\( \frac{1}{4} < x < 1 \)[/tex]
3. [tex]\( 1 < x < 4 \)[/tex]
4. [tex]\( x > 4 \)[/tex]
On calculating the values in these intervals, especially considering that the solution must satisfy the inequality [tex]\( 1 + \log_2 x - 2 \log_x 2 > 0 \)[/tex], we find that the inequality holds for [tex]\( 2 < x \)[/tex] and [tex]\(\frac{1}{4} < x < 1 \)[/tex].
So, the solution to [tex]\( 1 + \log_2 x - 2 \log_x 2 > 0 \)[/tex] is:
[tex]\[ (2 < x) \cup \left(\frac{1}{4} < x < 1\right) \][/tex]
This means [tex]\( x \)[/tex] must be in the intervals:
[tex]\[ x \in \left(2, \infty\right) \cup \left(\frac{1}{4}, 1\right) \][/tex]
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