Question

If the number of students taking online courses is increasing at a rate of 7% per year, How long will it take for the number of students taking online courses to double?​

Answer 1

Answer: D) about 10.24 years

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Work Shown:

The formula to use is $A = P(1+r)^t$, where,

A = final amount after t years

P = starting amount

r = growth rate

t = number of years

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The starting amount does not matter. Let's just make it some nice round number such as P = 100. This will double to A = 2*P = 2*100 = 200.

The growth rate in this case is r = 0.07

We want to solve for t

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$A = P(1+r)^t\\\\200 = 100(1+0.07)^t\\\\2 = (1.07)^t\\\\(1.07)^t = 2\\\\\log\left((1.07)^t\right) = \log(2)\\\\t*\log\left(1.07\right) = \log(2)\\\\t = \frac{\log(2)}{\log\left(1.07\right)}\\\\t \approx 10.2447683510588\\\\t \approx 10.24$

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Extra info:

$10.2447683510588 \ \text{years} = 10 \ \text{years} \ + \ 0.2447683510588 \ \text{years}\\\\10.2447683510588 \ \text{years} = 10 \ \text{years} \ + \ \left(0.2447683510588 \ \text{years}*\frac{12 \ \text{months}}{1 \ \text{year}}\right)\\\\10.2447683510588 \ \text{years} \approx 10 \ \text{years} \ + \ 2.9372202127056 \ \text{months}\\\\10.2447683510588 \ \text{years} \approx 10 \ \text{years}, \ 2.94 \ \text{months}$

and

$10.2447683510588 \ \text{years} = 10 \ \text{years} \ + \ 0.2447683510588 \ \text{years}\\\\10.2447683510588 \ \text{years} = 10 \ \text{years} \ + \ \left(0.2447683510588 \ \text{years}*\frac{365 \ \text{days}}{1 \ \text{year}}\right)\\\\10.2447683510588 \ \text{years} \approx 10 \ \text{years} \ + \ 89.340448136462 \ \text{days}\\\\10.2447683510588 \ \text{years} \approx 10 \ \text{years}, \ 89 \ \text{days}$

Answer 2

Answer:  D) 10.24 years

Step-by-step explanation:

Since the rate increases at the same rate every year, it is an exponential function.  Use the formula: $A=P(1+r)^t$   where

• A is the accrued amount (total population)
• P is the initial amount (starting population)
• r is the rate (convert percent to a decimal)
• t is the time (in years)

Given: A = 2P, P = P, r = 7% (0.07), t = unknown

$A=P(1+r)^t\\\\2P=P(1+0.07)^t\\\\2=1.07^t\qquad \qquad \text{(divide both sides by P)}\\\\ln(2)=ln(1.07)^t\quad \text{(apply ln to both sides)}\\\\ln(2)=t\ ln(1.07)\quad \text{(expand right side using log rules)}\\\\\dfrac{ln(2)}{ln(1.07)}=t \qquad \quad \text{(divide both sides by ln 1.07)}\\\\\\\large\boxed{10.24=t}$

It will take about 10.24 years for the number of students taking online courses to double.