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.)
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
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