Question 214807
{{{B(t)=12*e^((0.2*t))}}}
-----
substitute 1000 for B(t) to get:
{{{1000 = 12*e^((.2*t))}}}
take the natural log of both sides to get:
{{{ln(1000) = ln(12*e^(.2*t))}}}
-----
couple of general properties of logarithms are applicable.
the first is:
{{{ln (a*b) = ln(a) + ln(b)}}}
the second is:
{{{ln (a^b) = b*ln(a)}}}
-----
using the first one, we get:
{{{ln(12*e^(.2*t)) = ln(12) + ln(e^((.2*t)))}}}
-----
using the second one:
{{{ln(e^(.2*t)) = .2*t*ln(e)}}}
-----
putting this all together, your equation becomes:
{{{ln(1000) = ln(12) + .2*t*ln(e)}}}
-----
subtract {{{ln(12)}}} from both sides of this equation to get:
{{{ln(1000) - ln(12)  = .2*t*ln(e)}}}
-----
divide both sides of this equation by {{{.2*ln(e)}}} to get:
{{{(ln(1000) - ln(12)) / (ln(e)*.2) = t}}}
-----
this becomes:
{{{(6.907755279 - 2.48490665)/(1*(.2))}}}
-----
solve for t to get:
t = 22.11424315 hours.
-----
the population should increase to 1000 in 22.11424315 hours.
-----
substitute in original equation of:
{{{B(t)=12*e^((0.2*t))}}} to get:
{{{B(t)=12*e^((0.2*(22.11424315)))}}} which becomes:
B(t) = 1000 confirming t = 22.11424315 hours is a good value for t.
-----
Colony will increase to 1000 in 22.11424315 hours.
-----