Question

1. alex gets a 3% raise at the end of every year he works for a company. at the end of his fourth year, he is earning $50,648 dollars. find his starting salary, to the nearest dollar.
2. yeast is used in baking to make dough rise, which causes the volume of the dough to increase. the volume, in ml, of a certain dough can be modeled by the function $v(t) = 1500 \cdot 2^{\frac{t}{60}}$, where $t$ is the time in minutes since the dough was prepared.

a. how many minutes does it take the dough to double in volume?
b. find the hourly growth rate in the volume of the dough, as a percent. round to the nearest thousandth.

Explanation:

Problem 5

Step1: Define the formula for compound growth

The formula for compound growth is $A = P(1 + r)^n$, where $A$ is the final amount, $P$ is the principal (starting salary), $r$ is the annual growth rate (as a decimal), and $n$ is the number of years. Here, $r = 0.03$, $n = 4$, and $A = 50648$.

Step2: Solve for P

We need to solve $50648 = P(1 + 0.03)^4$ for $P$. First, calculate $(1.03)^4$. $(1.03)^4 \approx 1.12550881$. Then, $P = \frac{50648}{1.12550881} \approx 45000$.

Step1: Understand the doubling condition

We want to find $t$ when $V(t) = 2 \times 1500 = 3000$. The function is $V(t) = 1500 \cdot 2^{\frac{t}{60}}$.

Step2: Solve for t

Set $1500 \cdot 2^{\frac{t}{60}} = 3000$. Divide both sides by 1500: $2^{\frac{t}{60}} = 2$. Since the bases are equal, the exponents must be equal. So, $\frac{t}{60} = 1$. Multiply both sides by 60: $t = 60$.

Step1: Convert time to hours

We know the function is $V(t) = 1500 \cdot 2^{\frac{t}{60}}$. Let's find the volume after 1 hour (60 minutes) and then find the growth rate. First, find $V(60)$: $V(60) = 1500 \cdot 2^{\frac{60}{60}} = 1500 \cdot 2 = 3000$. Now, find the volume after 2 hours (120 minutes): $V(120) = 1500 \cdot 2^{\frac{120}{60}} = 1500 \cdot 4 = 6000$.

Step2: Calculate the growth rate

The growth rate formula for exponential growth is $A = P(1 + r)^n$, where $n = 1$ (1 hour), $P = 3000$, $A = 6000$. So, $6000 = 3000(1 + r)^1$. Solve for $r$: $1 + r = 2$, so $r = 1 = 100\%$? Wait, no, wait. Wait, the function is $V(t) = 1500 \cdot 2^{\frac{t}{60}}$. Let's rewrite it in terms of hours. Let $h = \frac{t}{60}$, so $t = 60h$. Then $V(h) = 1500 \cdot 2^{h}$. So the growth rate per hour is when $h$ increases by 1, the volume multiplies by 2. So the growth rate is $2 - 1 = 1 = 100\%$? Wait, but let's do it properly. The general form of exponential growth is $V(t) = V_0(1 + r)^t$ (here $t$ in hours). Our function is $V(t) = 1500 \cdot 2^{\frac{t}{60}} = 1500 \cdot (2^{\frac{1}{60}})^{t}$. Wait, no, earlier we converted $t$ to hours. Wait, maybe a better way: Let's find the growth factor per hour. Let $t_1 = 0$ hours (0 minutes), $V(0) = 1500$. $t_2 = 1$ hour (60 minutes), $V(60) = 3000$. The growth factor is $\frac{3000}{1500} = 2$. So the growth rate $r$ is such that $(1 + r)^1 = 2$, so $r = 1 = 100\%$? But that seems too high. Wait, no, the function is $V(t) = 1500 \cdot 2^{\frac{t}{60}}$. So when $t$ increases by 60 minutes (1 hour), the exponent increases by 1, so the volume doubles. So the hourly growth rate is 100%? Wait, but let's check with the formula for exponential growth. The function can be written as $V(t) = 1500 \cdot e^{kt}$. But we know it's a base-2 exponential. Alternatively, using the formula for growth rate: if $V(t) = V_0 \cdot a^t$ (t in hours), then the growth rate is $a - 1$. Here, when $t$ is in hours, $V(t) = 1500 \cdot 2^{t}$ (since $t$ hours is $60t$ minutes, so $\frac{60t}{60} = t$). So $a = 2$, so growth rate $r = 2 - 1 = 1 = 100\%$. Wait, but that's correct because it doubles every hour. Wait, but let's verify with the original function. At $t = 60$ minutes (1 hour), volume is 3000. At $t = 120$ minutes (2 hours), volume is 6000. So from 1 hour to 2 hours, it goes from 3000 to 6000, which is a 100% increase. So the hourly growth rate is 100.000%? Wait, but maybe I made a mistake. Wait, the function is $V(t) = 1500 \cdot 2^{\frac{t}{60}}$. Let's find the growth rate per hour. Let $t_1 = 0$, $V(0) = 1500$. $t_2 = 60$, $V(60) = 3000$. The growth rate from $t=0$ to $t=60$ (1 hour) is $\frac{3000 - 1500}{1500} = 1 = 100\%$. Then from $t=60$ to $t=120$ (another hour), $\frac{6000 - 3000}{3000} = 1 = 100\%$. So the hourly growth rate is 100.000%.

Answer:

The starting salary is $\$45000$.

Problem 6a