Question

rounding b to three decimal places, and using the form $p_n = p_0(b)^n$, the explicit formula for this problem is:
$p_n = 200(1.105)^n$
the question asks to predict the wolf population in 20 years, so $n = 20$, and results in $p_{20} = \square$ wolves.
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Explanation:

Step1: Identify the formula and values

We have the formula \( P_n = 200(1.105)^n \), and we need to find \( P_{20} \), so \( n = 20 \).

Step2: Substitute \( n = 20 \) into the formula

Calculate \( (1.105)^{20} \) first. Using a calculator, \( (1.105)^{20}\approx 6.7275 \) (this is an approximation, more precise calculation: \( 1.105^{20}=\mathrm{e}^{20\ln(1.105)}\approx\mathrm{e}^{20\times0.099758}\approx\mathrm{e}^{1.99516}\approx 7.353 \)? Wait, no, let's calculate directly: \( 1.105^{10}\approx 2.6533 \), then \( 1.105^{20}=(1.105^{10})^2\approx 2.6533^2\approx 7.040 \). Wait, maybe better to use calculator steps:

\( 1.105^1 = 1.105 \)

\( 1.105^2 = 1.105\times1.105 = 1.221025 \)

\( 1.105^3 = 1.221025\times1.105 \approx 1.349232625 \)

...

But using a calculator, \( 1.105^{20} \approx 7.353 \) (wait, no, let's check with a calculator: 1.105^20. Let's compute ln(1.105) = 0.099758, 200.099758 = 1.99516, e^1.99516 ≈ 7.35. Then 2007.35 ≈ 1470? Wait, no, maybe my initial approximation was wrong. Wait, let's use a calculator for 1.105^20:

Using a calculator, 1.105^20:

1.105^1 = 1.105

1.105^2 = 1.221025

1.105^3 = 1.221025 * 1.105 = 1.349232625

1.105^4 = 1.349232625 * 1.105 ≈ 1.490902051

1.105^5 ≈ 1.490902051 * 1.105 ≈ 1.647446766

1.105^6 ≈ 1.647446766 * 1.105 ≈ 1.820428676

1.105^7 ≈ 1.820428676 * 1.105 ≈ 2.011573687

1.105^8 ≈ 2.011573687 * 1.105 ≈ 2.222788924

1.105^9 ≈ 2.222788924 * 1.105 ≈ 2.456181761

1.105^10 ≈ 2.456181761 * 1.105 ≈ 2.714080846

1.105^11 ≈ 2.714080846 * 1.105 ≈ 2.999059335

1.105^12 ≈ 2.999059335 * 1.105 ≈ 3.313960565

1.105^13 ≈ 3.313960565 * 1.105 ≈ 3.661926424

1.105^14 ≈ 3.661926424 * 1.105 ≈ 4.046428698

1.105^15 ≈ 4.046428698 * 1.105 ≈ 4.471203711

1.105^16 ≈ 4.471203711 * 1.105 ≈ 4.940679101

1.105^17 ≈ 4.940679101 * 1.105 ≈ 5.459450407

1.105^18 ≈ 5.459450407 * 1.105 ≈ 6.032692699

1.105^19 ≈ 6.032692699 * 1.105 ≈ 6.666125432

1.105^20 ≈ 6.666125432 * 1.105 ≈ 7.366068602

Ah, so 1.105^20 ≈ 7.366068602

Then \( P_{20} = 200 \times 7.366068602 \approx 1473.2137204 \)

Wait, maybe my earlier steps were miscalculating. Let's use a calculator for 1.105^20:

Using a calculator, 1.105^20:

1.105^20 = e^(20 ln(1.105)) ≈ e^(20 0.099758) ≈ e^(1.99516) ≈ 7.353 (wait, no, ln(1.105) is approximately 0.099758, 20*0.099758=1.99516, e^1.99516 ≈ 7.35. But when calculating directly, we got ~7.366. The difference is due to more precise calculation of ln(1.105). Let's check ln(1.105):

ln(1.105) = 0.105 - 0.105^2/2 + 0.105^3/3 - 0.105^4/4 + ... (Taylor series around 0)

0.105 - 0.011025/2 + 0.001157625/3 - 0.000121550625/4 + ...

= 0.105 - 0.0055125 + 0.000385875 - 0.00003038765625 + ... ≈ 0.099847

Then 20*0.099847 = 1.99694

e^1.99694 ≈ e^2 / e^0.00306 ≈ 7.389056 / 1.003065 ≈ 7.366, which matches the direct calculation. So 1.105^20 ≈ 7.366

Then 200 * 7.366 ≈ 1473.2

So the population is approximately 1473 (or more precisely, 1473.21, so rounding to a whole number, 1473 or 1474? Wait, let's do the multiplication:

200 * 7.366068602 = 1473.2137204, so approximately 1473 wolves (if we round to the nearest whole number) or 1473.21 if we keep two decimals. But since it's a population, we usually round to the nearest whole number.

Wait, maybe the initial formula is correct, and we just need to compute 200*(1.105)^20.

Let's use a calculator for 1.105^20:

Using a calculator, 1.105^20 = 7.366068602

Then 200 * 7.366068602 = 1473.2137204

So \( P_{20} \approx 1473 \) (or 1473.21, depending on the requirement).

Answer:

\( P_{20} \approx \boxed{1473} \) (or if decimal is allowed, 1473.21)