Question

1. (4 ea) the assets, in dollars, for a financial firm can be approximated by the function $a(x)=2410 \cdot 2^{0.45x}$, where $x = 9$ corresponds to the year 2009. find the assets to the nearest dollar in each of the following years.

a. 2010
b. 2015
c. 2019

Explanation:

Step1: Find x for each year

For 2010: $x = 9 + (2010-2009) = 10$
For 2015: $x = 9 + (2015-2009) = 15$
For 2019: $x = 9 + (2019-2009) = 19$

Step2: Calculate 2010 assets

Substitute $x=10$ into $A(x)$:
$A(10) = 2410 \cdot 2^{0.45 \times 10} = 2410 \cdot 2^{4.5}$
$2^{4.5}=2^4 \cdot 2^{0.5}=16\sqrt{2}\approx16\times1.4142=22.6272$
$A(10)\approx2410\times22.6272\approx54531$

Step3: Calculate 2015 assets

Substitute $x=15$ into $A(x)$:
$A(15) = 2410 \cdot 2^{0.45 \times 15} = 2410 \cdot 2^{6.75}$
$2^{6.75}=2^6 \cdot 2^{0.75}=64\times2^{\frac{3}{4}}=64\times\sqrt[4]{8}\approx64\times1.6818=107.6352$
$A(15)\approx2410\times107.6352\approx259401$

Step4: Calculate 2019 assets

Substitute $x=19$ into $A(x)$:
$A(19) = 2410 \cdot 2^{0.45 \times 19} = 2410 \cdot 2^{8.55}$
$2^{8.55}=2^8 \cdot 2^{0.55}=256\times2^{0.55}\approx256\times1.4678=375.7568$
$A(19)\approx2410\times375.7568\approx905574$

Answer:

a. 2010: $\$54531$
b. 2015: $\$259401$
c. 2019: $\$905574$