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# 9th grade algebra help if you can!?

How long will it take for the population to reach 500?

### 5 Answers

- ?Lv 71 month agoFavorite Answer
225

Next year there are 225*1.15 = [left for you].

The next year there are [the above] * 1.15 = ?

The next, [the above] * 1.15 = ?

Do that till you reach the closest to 500.

Then, if it doesn't arrive exact, estimate the remaining.

So what do you get by this procedure?

Answer that, then we will show you the exact (algebra)

way to solve it.

Hopefully no one will spoil you the answer. That would be very irresponsible of them. And don't forget to vote me best answer for being the first to correctly walk you through and for not spoiling out the answer. That way it gives you a chance to work at it and to get good at it!

- lenpol7Lv 71 month ago
Use the Comoiund formula.

I(n) = I(o)[1 + r/100]^n

Where

I(n) = 500

I(o) = 225

r = 15%

n to be found.

Substitute

500 = 225[1 +15/100]^n

500/225 = [1 .15]^n

2.222.... = 1.15^n

Take natural logs (ln)

ln(2.222....) = ln(1.15)^n

ln(2.2222.....) = n*ln( 1.15)

n = ln(2.222...) / ln(1.15)

n = 0.7985.. / 0.13976

n = 5.71 years. (The Answer) .

- la consoleLv 71 month ago
v₀ = 225 ← initial value → 1 year later, the new value is:

v₁ = v₀ + (v₀ * 15 %) = v₀.(1 + 15 %) → 1 year more,

v₂ = v₁ + (v₁ * 15 %) = v₁.(1 + 15 %) = v₀.(1 + 15 %)² → 1 year later,

v₃ = v₂ + (v₂ * 15 %) = v₂.(1 + 15 %) = v₀.(1 + 15 %)³ → and you can generalize writing:

v(n) = v₀.(1 + 15 %)^(n) → where n represents the number of year

v(n) = v₀.(1 + 15 %)^(n) → you want to obtain 500

v₀.(1 + 15 %)^(n) = 500

(1 + 15 %)^(n) = 500/v₀ → recall: v₀ = 225

(1 + 15 %)^(n) = 500/225

(1 + 15 %)^(n) = 20/9

(1 + 0.15)^(n) = 20/9

1.15^(n) = 20/9

Ln[1.15^(n)] = Ln(20/9)

n.Ln(1.15) = Ln(20/9)

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

n ≈ 5.71

n = 6 years

- MorningfoxLv 71 month ago
What is 15% of 225? That's the increase for year 1.

Add: 225 + increase#1 = population at end of year 1.

Keep going for year 2, year 3, etc. Until you get a population of 500.

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- az_lenderLv 71 month ago
225*(1.15^x) = 500 =>

1.15^x = 500/225 = 20/9 =>

x log(1.15) = log (20/9) =>

x = log(20/9)/log(1.15) = 5.7 (years).

Seems about right -- the first year, the pop would increase by about 33, the next year by about 40, makes sense that several more years would be needed to get to 500.