exponetial growth question help

Algebra
Tutor: None Selected Time limit: 1 Day

Note: when solving for K, round to the four decimal place. 

A country's population in 1995 was 184 million. in 2002 it was 188 million. estimate the population in 2006 using exponential growth formula. round your answer to the nearest millions?

Aug 30th, 2015

Thank you for the opportunity to help you with your question!

184^n =188

introducing log

log (base184) 188=log 188/log184 =n

n= 1.004124

growth = X^ 1.004124

IF 1995 =n=0 and x being value of population in 1995

equation will be

184million^(1.004124(n))


in 2006 population will be

n= 11

population will be

=184^(1.004124*11)

= 3.6*10^25 million


Please let me know if you need any clarification. I'm always happy to answer your questions.
Aug 30th, 2015

sorry let me update it. It is wrong.

Aug 30th, 2015

184^(x^n)= 188 

if 1995 n=0

in 2002 n =7

184000000^(x^7)=188000000

introducing log

log (base184000000) 188000000=log 188000000/log184000000 =x^7 = 1.00113

x^7 =1.00113

x =7th root of 1.00113

x=1.0001614


formula is thus

184000000^(1.0001614^n)

in 2006 population will be

n=2006-1995=11

184000000^(1.0001614^11)= 184000000^1.001776 =190,326712.2 = 190 million


 

Aug 30th, 2015

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Aug 30th, 2015
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Aug 30th, 2015
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