To find the value of the table in 1980, we need to calculate \( v(0) \) since \( x = 0 \) corresponds to the year 1980.
The function for the value of the table is given by:
\[ v(x) = 750 \cdot (1.06)^{x} \]
Part 1: Value in 1980
Substituting \( x = 0 \):
\[ v(0) = 750 \cdot (1.06)^{0} = 750 \cdot 1 = 750 \]
Thus, the value of the table in 1980 was $750.00.
Part 2: Value in 2005
To find the value of the table in 2005, we need to calculate \( v(25) \) since 2005 is 25 years after 1980 (2005 - 1980 = 25).
\[ v(25) = 750 \cdot (1.06)^{25} \]
Calculating \( (1.06)^{25} \):
\[ (1.06)^{25} \approx 4.2908 \]
Now we can calculate \( v(25) \):
\[ v(25) = 750 \cdot 4.2908 \approx 3218.10 \]
Thus, the value of the table in 2005 was approximately $3218.10.
Part 3: Year when the table will reach double its 1980 value
Double the value of the table in 1980 is:
\[ 2 \times 750 = 1500 \]
To find the year when the table's value reaches $1500, we set up the equation:
\[ 750 \cdot (1.06)^{x} = 1500 \]
Dividing both sides by 750:
\[ (1.06)^{x} = 2 \]
Now, we can take the logarithm of both sides:
\[ \log((1.06)^{x}) = \log(2) \]
Using the power rule of logarithms:
\[ x \cdot \log(1.06) = \log(2) \]
Now, solve for \( x \):
\[ x = \frac{\log(2)}{\log(1.06)} \]
Calculating this using a calculator (approximately):
\[ \log(2) \approx 0.3010 \]
\[ \log(1.06) \approx 0.0253 \]
Thus,
\[ x \approx \frac{0.3010}{0.0253} \approx 11.88 \]
Since \( x \) is the number of years after 1980, we add this to 1980:
\[ 1980 + 11.88 \approx 1991.88 \]
So, the table is expected to reach double its 1980 value sometime in the year 1992 (rounding down since we are estimating the year).
Summary Results:
- a. The table was worth $750.00 in 1980.
- b. The value of the table in 2005 was approximately $3218.10.
- c. The table will reach double its 1980 value around the year 1992.
1 answer