exponential growth

Question

n a region, use of solar power has been growing exponentially. At the end of 2002, 1.5 terawatt-hours of solar power had been installed. This amount had risen to 32.4 terawatt-hours by the end of 2010. Assuming exponential growth, in what year will 6000 terawatt-hours of solar power be installed?

Answer 1

y = a·b^t

• y = terawatt hours
• a = starting number of terawatt hours = 1.5
• b = growth rate
• t = years since 2002

 

      y = 1.5·b^t

 

Plug in the ordered pair (2010, 32.4)

 

      32.4 = 1.5·b⁸

      32.4/1.5 = b⁸

      21.6 = b⁸

        ⁸√21.6 = b

      1.47 = b

 

      y = 1.5·(1.47)^t

 

      6000 = 1.5·(1.47)^t

      6000/1.5 = 1.47^t

      4000 = 1.47^t

      log(4000) = log(1.47^t)

      log(4000) = t·log(1.47)   [Log property: log(a^x) = x·log(a) for any base log]

      log(4000)/log(1.47) = t

 

Use your calculator to get the value of t.  You can use any base log.  Add t to 2002 to find the year.   Remember, no decimals in years.

Answer 2

From 2002 to 2010 (an 8 year time interval), it grew by a factor of 

 

32.4/1.5 = 21.6

 

Call the end of 2002 year zero, when the amount was 1.5 terawatt-hours.  Let t = number of years elapsed since the end of 2002.

 

A(t) = amount installed as function of number of years t.

 

A(t) = A(0) (21.6)^t/8

 

A(t) = 1.5 (21.6)^t/8 terawatt-hours 

 

Set 6000 = A(t) and solve for t.

 

6000 = 1.5 (21.6)^t/8

 

4000 = (21.6)^t/8

 

ln4000 = (t/8) ln21.6

 

t = 8 (ln4000 / ln21.6) years ≅ 21.59 years

 

Add that to the end of 2002 and that puts you at year 2023.59 to get to 6000 terawatt-hours, which means you get there shortly after mid 2024.