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# 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?

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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?
May 12, 2015

#2
+27342
+16

Continuing from zacismyname's analysis:

Take logs of both sides:

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

Because log(an) = 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

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May 12, 2015

#1
+981
+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.

May 12, 2015
#2
+27342
+16

Continuing from zacismyname's analysis:

Take logs of both sides:

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

Because log(an) = 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
#3
+95179
0

Just remember Zac,

If you need to find a power you can always resort to logs.

May 12, 2015
#4
+981
0

Yeah I haven't learnt that yet.

May 12, 2015