A business owner opens one store in town A. The equation p(x)=10,000(1.075)^t represents the anticipated profit after t years. The business owner opens a store in town B six months later and predicts the profit from that store to increase at the same rate. Assume that the initial profit from the store in town B is the same as the initial profit from the store in town A. At any time after both stores have opened, how does the profit from the store in town B compare with the profit from the store in town A?

Let's break down the problem step-by-step:

1. Profit Equation for Store in Town A:
   The profit for the store in town A is given by the equation:
   \[p_ A( t) = 10,000( 1.075) ^ t\]
   where \(t\) is the number of years since the store in town A opened.
2. Store in Town B Opens Six Months Later:
   The store in town B opens six months (0.5 years) after the store in town A. The initial profit for the store in town B is the same as the initial profit for the store in town A, which is 10,000 dollars.
3. Profit Equation for Store in Town B:
   Since the store in town B opens six months later, its profit can be represented similarly but with a time shift. Let \(t_ B\) be the number of years since the store in town B opened. Then, the profit for the store in town B is:
   \[p_ B( t_ B) = 10,000( 1.075) ^ { t_ B} \]
4. Relating Time for Both Stores:
   To compare profits at any time, we need to relate the time since the opening of each store. If the current time since the store in town A opened is \(t\) years, then the time since the store in town B opened is \(t - 0.5\) years (since it opened 0.5 years after the store in town A).
   Therefore, we can write:
   \[t_ B = t - 0.5\]
5. Comparing Profits:
   Substitute \(t_ B = t - 0.5\) into the profit equation for the store in town B:
   \[p_ B( t_ B) = 10,000( 1.075) ^ { t - 0.5} \]
   The profit for the store in town A at time \(t\) is:
   \[p_ A( t) = 10,000( 1.075) ^ t\]
   The profit for the store in town B at the same time is:
   \[p_ B( t - 0.5) = 10,000( 1.075) ^ { t - 0.5} \]
6. Conclusion:
   Since \(( 1.075) ^ { t - 0.5} \) is the profit growth factor for the store in town B, it is clear that the profit for the store in town B at time \(t\) is the same as the profit for the store in town A six months earlier. 
   Therefore, the profit from the store in town B will always be the profit from the store in town A six months later.

Supplemental Knowledge

Exponential functions are used to model situations where growth or decay happens at a constant percentage rate over time. The general form of an exponential growth function is:
\[p( t) = p_ 0 \cdot ( 1 + r) ^ t\]
where:

• \(p( t) \) is the profit at time \(t\),
• \(p_ 0\) is the initial profit,
• \(r\) is the growth rate,
• \(t\) is the time in years.

In this scenario, the profit from store A follows the equation:
\[p_ A( t) = 10,000( 1.075) ^ t\]
For store B, which opens six months later, we need to adjust for this delay. Since six months is half a year, if we denote the time since store B opened as \(t_ B\), then:
\[t_ B = t - 0.5\]
Thus, the profit function for store B would be:
\[p_ B( t) = 10,000( 1.075) ^ { t - 0.5} \]

Concepts to Actions

Imagine yourself as an entrepreneur operating two stores in different towns. After opening one first and seeing its growth steadily due to good business practices and customer satisfaction, six months later you open another with similar hopes for expansion.
Even though both stores begin with equal initial profit and progress at roughly equal rates, due to starting later, one will always lag behind another by about six months in terms of profits; at any given moment in time after both stores open, Store B's will equal or surpass that of Store A by approximately six months.

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