Question 1149203
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t = number of years after 1980
t = 0 represents the year 1980, t = 1 is 1981, and so on.


Plug t = 0 into the function to find the population in 1980


{{{N(t) = 217*e^(0.0102*t)}}}


{{{N(0) = 217*e^(0.0102*0)}}}


{{{N(0) = 217*e^(0)}}}


{{{N(0) = 217*1}}}


{{{N(0) = 217}}}


In 1980, there are 217 million people in that country.
Double this to get 2*217 = 434


The goal is to find the value of t such that N(t) = 434. 
We will use the natural logarithm function to help isolate t. 
Also, we'll use the log rules
<font color=blue>log rule 1: </font> {{{Ln(x^y) = y*Ln(x)}}}
<font color=blue>log rule 2: </font> {{{Ln(e) = 1}}}


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{{{N(t) = 217*e^(0.0102*t)}}}


{{{434 = 217*e^(0.0102*t)}}} Replace N(t) with 434


{{{217*e^(0.0102*t) = 434}}}


{{{(217*e^(0.0102*t))/217 = 434/217}}} Divide both sides by 217


{{{e^(0.0102*t) = 434/217}}} 


{{{e^(0.0102*t) = 2}}}


{{{Ln(e^(0.0102*t)) = Ln(2)}}} Apply natural logs to both sides


{{{0.0102*t*Ln(e) = Ln(2)}}} Use <font color=blue>log rule 1</font>


{{{0.0102*t*1 = Ln(2)}}} Use <font color=blue>log rule 2</font>


{{{0.0102*t = Ln(2)}}}


{{{(0.0102*t)/(0.0102) = Ln(2)/0.0102}}} Divide both sides by 0.0102


{{{t = Ln(2)/0.0102}}} 


{{{t = 67.9556059372496}}} Use a calculator. This is approximate


{{{t = "68.0"}}} Round to the nearest tenth (one decimal place)


{{{t = 68}}}


It will take about 68 for the population to double. 


Add this to 1980 to get 1980+68 = 2048


The population will double around the year <font color=red>2048</font>
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