Question 81409
Forumula for this type of problem would be {{{P=er^(t)}}} where P is the new amount, e is the original, r is the rate, and t is the time. You can create that formula by simply thinking about what the question is asking. If one year goes by, then the new amount of frogs is 18% greater, or {{{85 * 118%}}} OR {{{85 * 1.18}}}. If two years go by, then you'd get {{{85 * 1.18}}} after year one, and then multiply that by 1.18 AGAIN for the second year, so technically {{{85 * 1.18 * 1.18}}} Also known as {{{ 85 * 1.18^2}}} So....

a) the formula as a function would be... {{{f(t)=er^t}}}

b) plug in to find out the population after 3 years (t)
{{{f(3) = (85)(1.18^3)}}}
{{{f(3) = (85)(1.643032)}}}
{{{f(3) = 139.65772}}} Don't forget to round off, because you can't have .65772 of a frog, thats ridiculous. So 140 frogs after 3 years

c)plug in to find out the amount of f(t) = 600
{{{600 = (85)(1.18^t)}}}
{{{7.058823529 = 1.18^t}}}
{{{log 7.058823529 = t log 1.18}}}
{{{(log 7.058823529)/(log 1.18) = t}}}
{{{11.80729861 = t}}}
It would take 11.80729861 years to reach 600 frogs.