A logistic growth model for world population, f(x), in billions, x years after 1970 is f(x)= 12.57/(1+4.11e^-.026t). According to this model, the world population will be 9 billion in___.
+23252 f(x) = 9
9 = 12.57 / (1 + 4.11 e^( -.026t) )
9(1 + 4.11e^(-.026t) )= 12.57 Cross multiply
1 + 4.11e^(-.026t) = 12.57 / 9 Divide by 9. Feel free to calculate and use decimals from here.
4.11e^(-.026t) = 12.57 / 9 - 1 Subtract 1.
e^(-.026t) = ( 12.57 / 9 - 1 ) / 4.11 Divide by 4.11.
ln( e^(-.026t) ) = ln( (12.57 / 9 - 1) / 4.11 ) Since the variable is in the exponent, find the ln of both sides; also, use log instead, if you wish.
-.026t x ln( e ) = ln( (12.57 / 9 - 1 ) / 4.11 ) The exponent comes out as a multiplier.
-.026t = ln( (12.57 / 9 - 1) / 4.11 ) The ln(e) = 1; it will disappear in multiplication.
t = ( ln( ( 12.57 / 9 - 1 ) / 4.11 ) ) / -.026 Divide by -.026.
Add this answer to 1970.
+23252 f(x) = 9
9 = 12.57 / (1 + 4.11 e^( -.026t) )
9(1 + 4.11e^(-.026t) )= 12.57 Cross multiply
1 + 4.11e^(-.026t) = 12.57 / 9 Divide by 9. Feel free to calculate and use decimals from here.
4.11e^(-.026t) = 12.57 / 9 - 1 Subtract 1.
e^(-.026t) = ( 12.57 / 9 - 1 ) / 4.11 Divide by 4.11.
ln( e^(-.026t) ) = ln( (12.57 / 9 - 1) / 4.11 ) Since the variable is in the exponent, find the ln of both sides; also, use log instead, if you wish.
-.026t x ln( e ) = ln( (12.57 / 9 - 1 ) / 4.11 ) The exponent comes out as a multiplier.
-.026t = ln( (12.57 / 9 - 1) / 4.11 ) The ln(e) = 1; it will disappear in multiplication.
t = ( ln( ( 12.57 / 9 - 1 ) / 4.11 ) ) / -.026 Divide by -.026.
Add this answer to 1970.
