Answer :
To solve for the positive and negative intervals of the function [tex]\( r(x) = 2 \cdot (1.5)^x - 3 \)[/tex], we need to determine the intervals where the function is greater than zero (positive) and where the function is less than zero (negative).
Step-by-Step Solution:
1. Rewrite the function: Given [tex]\( r(x) = 2 \cdot (1.5)^x - 3 \)[/tex].
2. Set the function to inequalities to find the intervals:
- For positive intervals: [tex]\( r(x) > 0 \)[/tex]
[tex]\[ 2 \cdot (1.5)^x - 3 > 0 \][/tex]
Simplifying, we get:
[tex]\[ 2 \cdot (1.5)^x > 3 \][/tex]
Dividing both sides by 2:
[tex]\[ (1.5)^x > \frac{3}{2} \][/tex]
- For negative intervals: [tex]\( r(x) < 0 \)[/tex]
[tex]\[ 2 \cdot (1.5)^x - 3 < 0 \][/tex]
Simplifying, we get:
[tex]\[ 2 \cdot (1.5)^x < 3 \][/tex]
Dividing both sides by 2:
[tex]\[ (1.5)^x < \frac{3}{2} \][/tex]
3. Solving the inequalities:
- Inequality [tex]\( (1.5)^x > \frac{3}{2} \)[/tex]:
[tex]\[ x > \log_{1.5}\left(\frac{3}{2}\right) \][/tex]
Converting to natural log, knowing [tex]\( a \log_b(c) = \frac{\log_a(c)}{\log_a(b)} \)[/tex]:
[tex]\[ x > \frac{\log(1.5)}{\log(3/2)} \][/tex]
Numerically solving this, we get:
[tex]\[ x > 2.46630346237643 \cdot \log(3/2) \][/tex]
- Inequality [tex]\( (1.5)^x < \frac{3}{2} \)[/tex]:
[tex]\[ x < \log_{1.5}\left(\frac{3}{2}\right) \][/tex]
Converting to natural log, knowing [tex]\( a \log_b(c) = \frac{\log_a(c)}{\log_a(b)} \)[/tex]:
[tex]\[ x < \frac{\log(1.5)}{\log(3/2)} \][/tex]
Numerically solving this, we get:
[tex]\[ x < 2.46630346237643 \cdot \log(3/2) \][/tex]
4. Determine the intervals:
- Positive interval:
[tex]\[ x > 2.46630346237643 \cdot \log(3/2) \][/tex]
- Negative interval:
[tex]\[ x < 2.46630346237643 \cdot \log(3/2) \][/tex]
So, the positive interval for the function [tex]\( 2 \cdot (1.5)^x - 3 \)[/tex] is [tex]\( x > 2.46630346237643 \cdot \log(3/2) \)[/tex] and the negative interval is [tex]\( x < 2.46630346237643 \cdot \log(3/2) \)[/tex].
These intervals indicate the values of [tex]\( x \)[/tex] on which the function [tex]\( r(x) \)[/tex] is positive and negative, respectively.
Step-by-Step Solution:
1. Rewrite the function: Given [tex]\( r(x) = 2 \cdot (1.5)^x - 3 \)[/tex].
2. Set the function to inequalities to find the intervals:
- For positive intervals: [tex]\( r(x) > 0 \)[/tex]
[tex]\[ 2 \cdot (1.5)^x - 3 > 0 \][/tex]
Simplifying, we get:
[tex]\[ 2 \cdot (1.5)^x > 3 \][/tex]
Dividing both sides by 2:
[tex]\[ (1.5)^x > \frac{3}{2} \][/tex]
- For negative intervals: [tex]\( r(x) < 0 \)[/tex]
[tex]\[ 2 \cdot (1.5)^x - 3 < 0 \][/tex]
Simplifying, we get:
[tex]\[ 2 \cdot (1.5)^x < 3 \][/tex]
Dividing both sides by 2:
[tex]\[ (1.5)^x < \frac{3}{2} \][/tex]
3. Solving the inequalities:
- Inequality [tex]\( (1.5)^x > \frac{3}{2} \)[/tex]:
[tex]\[ x > \log_{1.5}\left(\frac{3}{2}\right) \][/tex]
Converting to natural log, knowing [tex]\( a \log_b(c) = \frac{\log_a(c)}{\log_a(b)} \)[/tex]:
[tex]\[ x > \frac{\log(1.5)}{\log(3/2)} \][/tex]
Numerically solving this, we get:
[tex]\[ x > 2.46630346237643 \cdot \log(3/2) \][/tex]
- Inequality [tex]\( (1.5)^x < \frac{3}{2} \)[/tex]:
[tex]\[ x < \log_{1.5}\left(\frac{3}{2}\right) \][/tex]
Converting to natural log, knowing [tex]\( a \log_b(c) = \frac{\log_a(c)}{\log_a(b)} \)[/tex]:
[tex]\[ x < \frac{\log(1.5)}{\log(3/2)} \][/tex]
Numerically solving this, we get:
[tex]\[ x < 2.46630346237643 \cdot \log(3/2) \][/tex]
4. Determine the intervals:
- Positive interval:
[tex]\[ x > 2.46630346237643 \cdot \log(3/2) \][/tex]
- Negative interval:
[tex]\[ x < 2.46630346237643 \cdot \log(3/2) \][/tex]
So, the positive interval for the function [tex]\( 2 \cdot (1.5)^x - 3 \)[/tex] is [tex]\( x > 2.46630346237643 \cdot \log(3/2) \)[/tex] and the negative interval is [tex]\( x < 2.46630346237643 \cdot \log(3/2) \)[/tex].
These intervals indicate the values of [tex]\( x \)[/tex] on which the function [tex]\( r(x) \)[/tex] is positive and negative, respectively.
