There are 225 trout in a lake. The population is increasing of 15% per year. At this rate when will the population reach 500 trout?

Guest May 12, 2015

#2**+16 **

Continuing from zacismyname's analysis:

Take logs of both sides:

log(1.15^{n}) = log(20/9)

Because log(a^{n}) = n*log(a) we can write

n*log(1.15) = log(20/9)

so

n = log(20/9)/log(1.15)

$${\mathtt{n}} = {\frac{{log}_{10}\left({\frac{{\mathtt{20}}}{{\mathtt{9}}}}\right)}{{log}_{10}\left({\mathtt{1.15}}\right)}} \Rightarrow {\mathtt{n}} = {\mathtt{5.713\: \!341\: \!433\: \!638\: \!366\: \!8}}$$

n ≈ 5.7 years

.

Alan
May 12, 2015

#1**+13 **

Trout Population = starting population*(1 + rate)^number of years

$$P = SP\times{(1+0.15)^n$$

$$500 = 225\times{(1+0.15)^n$$

$$1.15^n = \frac{20}{9}$$

Now I'm not sure how you find n. It is somewhere between 5 and 6.

zacismyname
May 12, 2015

#2**+16 **

Best Answer

Continuing from zacismyname's analysis:

Take logs of both sides:

log(1.15^{n}) = log(20/9)

Because log(a^{n}) = n*log(a) we can write

n*log(1.15) = log(20/9)

so

n = log(20/9)/log(1.15)

$${\mathtt{n}} = {\frac{{log}_{10}\left({\frac{{\mathtt{20}}}{{\mathtt{9}}}}\right)}{{log}_{10}\left({\mathtt{1.15}}\right)}} \Rightarrow {\mathtt{n}} = {\mathtt{5.713\: \!341\: \!433\: \!638\: \!366\: \!8}}$$

n ≈ 5.7 years

.

Alan
May 12, 2015