If a quantity grows by 7% per year, how long will it take to double?

To calculate the time it takes for a quantity to double with an exponential growth of 7% per year, we’ll use the following formula:

A(t) = A₀(1 + r)^t

Where:

- A(t) is the amount at time t

- A₀ is the initial amount

- r is the annual growth rate (as a decimal)

- t is the time in years

We want to find the value of t when the amount doubles, so we can set A(t) to 2A₀:

2A₀ = A₀(1 + r)^t

Now, we can divide both sides by A₀ and plug in the growth rate of 7% = 0.07:

2 = (1 + 0.07)^t

To solve for t, we’ll use the logarithm. We can take the natural logarithm of both sides:

ln (2) = ln ((1+0.07)^t)

Next, we will use the property of logarithms ln (a^b) = bln (a):

ln (2) = tln (1 + 0.07)

Now, we can solve for t:

$$t = \frac{\ln(2)}{\ln(1+0.07)}$$

Calculating the value:

$$t \approx \frac{0.693}{0.068}$$

t ≈ 10.24

So, it will take approximately 10.24 years for a quantity to double with a 7% annual growth rate.