+0  
 
0
1
1639
2
avatar

The exponential function 

N = 3.93 × 1.34 ^d gives the approximate U.S. population, in millions, d decades after 1790. (The formula is valid only up to 1860.)

 

 

Find a formula that gives the U.S. population c centuries after 1790. (Assume that the original formula is valid over several centuries.)

 Nov 7, 2015

Best Answer 

 #2
avatar+118587 
+5

The exponential function 

N = 3.93 × 1.34 ^d gives the approximate U.S. population, in millions, d decades after 1790. (The formula is valid only up to 1860.)

 

 

Find a formula that gives the U.S. population c centuries after 1790. (Assume that the original formula is valid over several centuries.)

 

\(3.93 × 1.34 ^d=3.93\times a^{0.1d}\\ 1.34 ^d=a^{0.1d}\\ log(1.34 ^d)=log(a^{0.1d})\\ d*log(1.34)=0.1d*log(a)\\ log(1.34)=0.1*log(a)\\ 10log(1.34)=log(a)\\ log(a)=10log(1.34)\\ 10^{log(a)}=10^{10*log(1.34)}\\ a=10^{10*log(1.34)}\\ a\approx 18.6659\qquad (4dp)\)

 

N = 3.93 × 1.34 ^d

N = 3.93 × 18.6659 ^C                         C for century

 

check after 1/2 cnetury

N = 3.93 × 1.34 ^5                 =16.979

N = 3.93 × 18.6659 ^0.5       = 16.979

after 5 centuries

N = 3.93 × 1.34 ^50              = 8 904 970

N = 3.93 × 18.6659 ^5          = 8 905 067         difference due to rounding

 Nov 8, 2015
 #1
avatar+104 
+5

n=(((393*50^(-d-1))*67^d)/2).

 Nov 7, 2015
 #2
avatar+118587 
+5
Best Answer

The exponential function 

N = 3.93 × 1.34 ^d gives the approximate U.S. population, in millions, d decades after 1790. (The formula is valid only up to 1860.)

 

 

Find a formula that gives the U.S. population c centuries after 1790. (Assume that the original formula is valid over several centuries.)

 

\(3.93 × 1.34 ^d=3.93\times a^{0.1d}\\ 1.34 ^d=a^{0.1d}\\ log(1.34 ^d)=log(a^{0.1d})\\ d*log(1.34)=0.1d*log(a)\\ log(1.34)=0.1*log(a)\\ 10log(1.34)=log(a)\\ log(a)=10log(1.34)\\ 10^{log(a)}=10^{10*log(1.34)}\\ a=10^{10*log(1.34)}\\ a\approx 18.6659\qquad (4dp)\)

 

N = 3.93 × 1.34 ^d

N = 3.93 × 18.6659 ^C                         C for century

 

check after 1/2 cnetury

N = 3.93 × 1.34 ^5                 =16.979

N = 3.93 × 18.6659 ^0.5       = 16.979

after 5 centuries

N = 3.93 × 1.34 ^50              = 8 904 970

N = 3.93 × 18.6659 ^5          = 8 905 067         difference due to rounding

Melody Nov 8, 2015

3 Online Users

avatar
avatar