The formula $$ S=C(1+r)^{t} $$ models inflation, where $$ C= $$ the value today, $$ r= $$ the annual inflation rate (in decimal form), and $$ S= $$ the inflated value $$ t $$ years from now. If the inflation rate is $$ 3 \% $$, how much will a house now worth $$ \$ 153,000 $$ be worth in 7 years? Round your answer to the nearest dollar. The house will be worth $$ \$ \square $$. (Round to the nearest dollar as needed.)

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To find the inflated value of the house in 7 years, we can use the given formula for inflation:

$$ S = C(1 + r)^t $$

Given:
- $$ C = \$153,000 $$ (the value today)
- $$ r = 3\% = 0.03 $$ (the annual inflation rate in decimal form)
- $$ t = 7 $$ years

Substitute the given values into the formula:

$$ S = 153,000(1 + 0.03)^7 $$

Now, we can calculate the inflated value of the house in 7 years.
Calculate the value by following steps:
- step0: Calculate:
  $$153000\left(1+0.03\right)^{7}$$
- step1: Add the numbers:
  $$153000\times 1.03^{7}$$
- step2: Convert the expressions:
  $$153000\left(\frac{103}{100}\right)^{7}$$
- step3: Simplify:
  $$153000\times \frac{103^{7}}{100^{7}}$$
- step4: Rewrite the expression:
  $$1000\times 153\times \frac{103^{7}}{100^{7}}$$
- step5: Rewrite the expression:
  $$10^{3}\times 153\times \frac{103^{7}}{10^{14}}$$
- step6: Reduce the numbers:
  $$153\times \frac{103^{7}}{10^{11}}$$
- step7: Multiply:
  $$\frac{153\times 103^{7}}{10^{11}}$$
The house will be worth approximately $188,170.70 in 7 years. Rounding to the nearest dollar, the house will be worth $188,171.
The house will be worth \$188,171 in 7 years.

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