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Simplify the expression [tex]$(5^2)^3$[/tex].

1. Apply the power of a power property: [tex]$a^{m^n} = a^{m \cdot n}$[/tex].
2. Simplify: [tex]$(5^2)^3 = 5^{2 \cdot 3} = 5^6$[/tex].
3. Calculate: [tex]$5^6 = 15,625$[/tex].

Thus, [tex]$(5^2)^3 = 15,625$[/tex].

Answer :

To simplify the expression [tex](5^2)^3[/tex], we can apply the power of a power rule from the laws of exponents. This rule states that [tex](a^m)^n = a^{m \cdot n}[/tex], where [tex]a[/tex] is the base and [tex]m[/tex] and [tex]n[/tex] are the exponents.

Step-by-Step Explanation:

  1. Identify the base and the exponents in the expression. In this case, the base is [tex]5[/tex], the inner exponent is [tex]2[/tex], and the outer exponent is [tex]3[/tex].

  2. Apply the power of a power rule:
    [tex](5^2)^3 = 5^{2 \cdot 3}[/tex].

  3. Multiply the exponents:
    [tex]2 \times 3 = 6[/tex].

  4. Simplify the expression:
    [tex]5^{2 \cdot 3} = 5^6[/tex].

  5. Compute [tex]5^6[/tex]:

    Let's break it down:

    • [tex]5 \times 5 = 25[/tex]
    • [tex]25 \times 5 = 125[/tex]
    • [tex]125 \times 5 = 625[/tex]
    • [tex]625 \times 5 = 3125[/tex]
    • [tex]3125 \times 5 = 15625[/tex]

    Thus, [tex]5^6 = 15625[/tex].

Real-World Application: Population Growth Simulation

Now, let's connect this to a real-world scenario where the dog shelter's population doubles every month. This situation is modeled by exponential growth.

Procedure:

  1. If the initial population is [tex]P_0[/tex], the population doubles every month, meaning the growth factor is [tex]2[/tex].

  2. Population after [tex]n[/tex] months: [tex]P_n = P_0 \times 2^n[/tex].

  3. Calculate the population over a six-month period. If [tex]P_0 = 1[/tex] for simplicity:

    • After 1 month: [tex]P_1 = 1 \times 2^1 = 2[/tex]
    • After 2 months: [tex]P_2 = 1 \times 2^2 = 4[/tex]
    • After 3 months: [tex]P_3 = 1 \times 2^3 = 8[/tex]
    • After 4 months: [tex]P_4 = 1 \times 2^4 = 16[/tex]
    • After 5 months: [tex]P_5 = 1 \times 2^5 = 32[/tex]
    • After 6 months: [tex]P_6 = 1 \times 2^6 = 64[/tex]

Discussion:

Exponential growth is characterized by a constant doubling or rate of increase applied consistently, leading to rapid escalation. In contrast, linear growth involves a constant addition of the same amount each period, creating a straight-line increase. The dog shelter example highlights exponential growth due to the doubling nature of the population, demonstrating how calculations like [tex](5^2)^3[/tex] can easily relate to applications in real-world scenarios, like modeling populations over time.

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Rewritten by : Jeany

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