Approximately what number of clicks would get you to 1.5 ft in size? 18= 1 (1.5)* x = 1n( 18) In(15) would 4: What if we started with the house having a size of 1.5 ft.? If a click to go smaller reduces the size by 50%, what formula could we use that is well connected to the first one? Converted to inches, we have:
Approximately what number of clicks would get you to 1.5 ft in size? 18= 1 (1.5)* x = 1n( 18) In(15) would 4: What if we started with the house having a size of 1.5 ft.? If a click to go smaller reduces the size by 50%, what formula could we use that is well connected to the first one? Converted to inches, we have:
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![### Exponential Growth and Reduction:
When working with exponential formulas to determine the size of an object over a series of actions (clicks), we can use specific functions to illustrate this process.
Let’s consider the function:
\[ f(x) = 1 \cdot (1.5)^x \]
This function represents the growth of a house in feet after each click, where \(x\) is the number of clicks.
#### Example Calculation:
**How big is the house after 5 clicks?**
\[ f(5) = 1 \cdot (1.5)^5 \]
\[ = 1 \cdot 7.59375 \]
\[ = 7.59375 \, \text{inches} \]
To find out how many clicks are needed to reach a specific size, for instance, 1.5 feet, we reformulate the equation as follows:
\[ 18 = 1 \cdot (1.5)^x \]
\[ x = \frac{\ln(18)}{\ln(1.5)} \]
This equation uses the natural logarithm (ln) to solve for \(x\).
### Practical Problem:
**Task 4:**
If we start with a house having a size of 1.5 feet and a click reduces the size by 50%, we should use a similar exponential formula to determine the new size. When converted to inches, we have the following formula:
\[ g(x) = 18 \cdot (0.5)^x \]
This new function represents the reduction in size over clicks.
By understanding these exponential functions, students can analyze both growth and reduction scenarios dynamically.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F71e9498b-aba9-407d-bec7-742f7e9d7c6e%2F7b8e5b61-b1f2-4a9d-b3c1-20631fa7bd22%2Flgkb0xb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Exponential Growth and Reduction:
When working with exponential formulas to determine the size of an object over a series of actions (clicks), we can use specific functions to illustrate this process.
Let’s consider the function:
\[ f(x) = 1 \cdot (1.5)^x \]
This function represents the growth of a house in feet after each click, where \(x\) is the number of clicks.
#### Example Calculation:
**How big is the house after 5 clicks?**
\[ f(5) = 1 \cdot (1.5)^5 \]
\[ = 1 \cdot 7.59375 \]
\[ = 7.59375 \, \text{inches} \]
To find out how many clicks are needed to reach a specific size, for instance, 1.5 feet, we reformulate the equation as follows:
\[ 18 = 1 \cdot (1.5)^x \]
\[ x = \frac{\ln(18)}{\ln(1.5)} \]
This equation uses the natural logarithm (ln) to solve for \(x\).
### Practical Problem:
**Task 4:**
If we start with a house having a size of 1.5 feet and a click reduces the size by 50%, we should use a similar exponential formula to determine the new size. When converted to inches, we have the following formula:
\[ g(x) = 18 \cdot (0.5)^x \]
This new function represents the reduction in size over clicks.
By understanding these exponential functions, students can analyze both growth and reduction scenarios dynamically.
![**Exploring Exponential Growth and Decay in Measurements**
**Understanding the Problem Statement:**
*What if we started with the house having a size of 1.5 ft. If a click to go smaller decreases the size by 50%, what formula could we use that is well connected to the first one? Convert to have:*
\[ f(x) = 18(.5)^x \]
**Objective:**
Determine how many clicks it would take to get back to an inch.
**Step-by-Step Solution:**
1. **Understanding the Exponential Function:**
The given formula \( f(x) = 18(.5)^x \) represents an exponential decay where:
- 18 is the initial size in inches.
- \( .5 \) is the decay factor per click.
- \( x \) is the number of clicks.
2. **Converting Feet to Inches:**
Since there are 12 inches in a foot, convert 1.5 ft to inches.
\[ 1.5 \text{ ft} = 1.5 \times 12 = 18 \text{ inches} \]
This matches our initial size in the formula.
3. **Finding the Desired Size:**
Now, we need to determine the number of clicks it would take for the size to be 1 inch.
Set \( f(x) \) to 1:
\[ 1 = 18(.5)^x \]
4. **Solving for \( x \):**
Isolate the exponential term:
\[ (.5)^x = \frac{1}{18} \]
Take the logarithm of both sides to solve for \( x \):
\[ x \log(0.5) = \log\left(\frac{1}{18}\right) \]
\[ x = \frac{\log\left(\frac{1}{18}\right)}{\log(0.5)} \]
Calculate the values:
\[ \log\left(\frac{1}{18}\right) \approx \log(0.0556) \approx -1.254 \]
\[ \log(0.5) \approx -0.301 \]
Plug these values into the equation:
\[ x \approx \frac{-1.254}{-0.301} \approx 4.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F71e9498b-aba9-407d-bec7-742f7e9d7c6e%2F7b8e5b61-b1f2-4a9d-b3c1-20631fa7bd22%2F9d2a8km_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Exploring Exponential Growth and Decay in Measurements**
**Understanding the Problem Statement:**
*What if we started with the house having a size of 1.5 ft. If a click to go smaller decreases the size by 50%, what formula could we use that is well connected to the first one? Convert to have:*
\[ f(x) = 18(.5)^x \]
**Objective:**
Determine how many clicks it would take to get back to an inch.
**Step-by-Step Solution:**
1. **Understanding the Exponential Function:**
The given formula \( f(x) = 18(.5)^x \) represents an exponential decay where:
- 18 is the initial size in inches.
- \( .5 \) is the decay factor per click.
- \( x \) is the number of clicks.
2. **Converting Feet to Inches:**
Since there are 12 inches in a foot, convert 1.5 ft to inches.
\[ 1.5 \text{ ft} = 1.5 \times 12 = 18 \text{ inches} \]
This matches our initial size in the formula.
3. **Finding the Desired Size:**
Now, we need to determine the number of clicks it would take for the size to be 1 inch.
Set \( f(x) \) to 1:
\[ 1 = 18(.5)^x \]
4. **Solving for \( x \):**
Isolate the exponential term:
\[ (.5)^x = \frac{1}{18} \]
Take the logarithm of both sides to solve for \( x \):
\[ x \log(0.5) = \log\left(\frac{1}{18}\right) \]
\[ x = \frac{\log\left(\frac{1}{18}\right)}{\log(0.5)} \]
Calculate the values:
\[ \log\left(\frac{1}{18}\right) \approx \log(0.0556) \approx -1.254 \]
\[ \log(0.5) \approx -0.301 \]
Plug these values into the equation:
\[ x \approx \frac{-1.254}{-0.301} \approx 4.
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