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:
Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 miles away.
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experiments to set up the devices.
![### 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)
![**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)
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