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The population, [tex]p[/tex], of a town after [tex]t[/tex] years is represented using the equation [tex]p=10000(1.04)^{-t}[/tex]. Which of the following is an equivalent expression?

A. [tex]p=10000\left(\frac{1}{25}\right)^t[/tex]
B. [tex]p=10000\left(\frac{25}{26}\right)^t[/tex]
C. [tex]p=10000\left(\frac{26}{25}\right)^t[/tex]
D. [tex]p=10000\left(\frac{25}{1}\right)^t[/tex]

Answer :

To find an equivalent expression for the given population equation [tex]\( p = 10000(1.04)^{-t} \)[/tex], let's break down the process step-by-step.

1. Understand the Expression: The original expression given is [tex]\( p = 10000(1.04)^{-t} \)[/tex]. Here, [tex]\( 1.04 \)[/tex] is the base, and the exponent is [tex]\(-t\)[/tex].

2. Rewrite [tex]\( 1.04 \)[/tex]: We want to express [tex]\( 1.04 \)[/tex] in a form that could make the expression clearer. Observe that [tex]\( 1.04 \)[/tex] can be represented as a fraction:
[tex]\( 1.04 = \frac{26}{25} \)[/tex].
This is because [tex]\( 1.04 = 1 + 0.04 = \frac{25}{25} + \frac{1}{25} = \frac{26}{25} \)[/tex].

3. Apply the Negative Exponent Rule: The expression [tex]\( (1.04)^{-t} \)[/tex] becomes [tex]\( \left(\frac{26}{25}\right)^{-t} \)[/tex].
Applying the negative exponent rule, which is [tex]\( a^{-b} = \left(\frac{1}{a}\right)^b \)[/tex], we get:
[tex]\(\left(\frac{26}{25}\right)^{-t} = \left(\frac{1}{\frac{26}{25}}\right)^t = \left(\frac{25}{26}\right)^t\)[/tex].

4. Update the Expression: Substitute back into the original equation with this new base:
[tex]\[
p = 10000 \times \left(\frac{25}{26}\right)^t
\][/tex]

Therefore, the equivalent expression is:
[tex]\[ p = 10000\left(\frac{25}{26}\right)^t \][/tex]

This matches the second option from the multiple-choice answers.

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