6. A painting was worth $$ \$ 500 $$ when it was first purchased in 1992 . In 2020 , it was worth $$ \$ 6750 $$. (a) Write an exponential function that could be used to model the value of the painting, $$ y $$ (Round 'b" to 4 decimal places) (b) What is the rate of growth? (c) If the value of the paining continues to grow at the same rato, how much will it be worth in '2035?

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**(a)** Let $$ t $$ be the year. We express the value with an exponential function of the form
$$
y = 500 \cdot b^{(t-1992)}.
$$
We know that in 2020 ($$ t = 2020 $$), the painting is worth $$ \$6750 $$. Thus,
$$
6750 = 500 \cdot b^{2020-1992} = 500 \cdot b^{28}.
$$
Dividing both sides by 500 gives
$$
b^{28} = \frac{6750}{500} = 13.5.
$$
Taking the 28th root of both sides (or equivalently using logarithms),
$$
b = 13.5^{\frac{1}{28}}.
$$
Evaluating this expression and rounding to 4 decimal places,
$$
b \approx 1.0974.
$$
Thus, the exponential function is
$$
y = 500 \cdot (1.0974)^{(t-1992)}.
$$

**(b)** The rate of growth per year is given by
$$
\text{Growth rate} = b - 1 \approx 1.0974 - 1 = 0.0974,
$$
or about $$9.74\%$$ per year.

**(c)** To find the value of the painting in $$2035$$, note that
$$
\text{Number of years since 1992} = 2035 - 1992 = 43.
$$
Substitute $$ t-1992 = 43 $$ into the function:
$$
y = 500 \cdot (1.0974)^{43}.
$$
Evaluating the exponent,
$$
(1.0974)^{43} \approx e^{43 \ln(1.0974)}.
$$
Since $$\ln(1.0974) \approx 0.09295$$, we have
$$
43 \ln(1.0974) \approx 43 \times 0.09295 \approx 3.9969,
$$
and thus,
$$
(1.0974)^{43} \approx e^{3.9969} \approx 54.401.
$$
Therefore,
$$
y \approx 500 \cdot 54.401 \approx 27200.5.
$$
Rounding to the nearest dollar, the painting will be worth approximately $$\$27,\!200$$ in 2035.
**(a)** The exponential function modeling the painting's value is: $$ y = 500 \cdot (1.0974)^{(t-1992)} $$ **(b)** The rate of growth is approximately $$9.74\%$$ per year. **(c)** In 2035, the painting will be worth about \$27,200.

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