Question

from january 2014 to january 2019 the number ( n(t) ) of wikipedia articles(^1) was approximated by ( n(t) = n_0 , e^{0.054t} ) where ( t ) is the number of years since january 1, 2014.
the doubling time for the number of wikipedia articles during this period is:
note: round your answer to two decimal places.

( \boxed{} ) years

(^1)en.wikipedia.org/wiki/wikipedia:size_of_wikipedia
annual_growth_rate_for_the_english_wikipedia, accessed march 4, 2020.

Explanation:

Step1: Set up the doubling time equation

We know that for exponential growth \( N(t) = N_0 e^{rt} \), when the number doubles, \( N(t) = 2N_0 \). So we set \( 2N_0 = N_0 e^{0.054t} \).

Step2: Solve for \( t \)

Divide both sides by \( N_0 \): \( 2 = e^{0.054t} \).
Take the natural logarithm of both sides: \( \ln(2) = \ln(e^{0.054t}) \).
Using the property \( \ln(e^x)=x \), we get \( \ln(2) = 0.054t \).
Then solve for \( t \): \( t=\frac{\ln(2)}{0.054} \).
Calculate \( \ln(2)\approx0.6931 \), so \( t = \frac{0.6931}{0.054}\approx12.835 \).
Round to two decimal places: \( t\approx12.84 \).

Answer:

12.84