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powers: Exercises 16,18,20,22,24,26 p. 214

\(3^4=81 \)

\(3^{-2} = \frac{1}{9} \)

\(81^{1/2} = \left( 9^2 \right)^{1/2} = 9^{2 \times 1/2} = 9^1=9 \)

\(2^3 \times 2^5 = 2^{3+5}=2^8=256 \)

\(6^2 \times 6^{-2} = 6^{2-2} = 6^0 = 1\)

\(3^3+2^3 = 27+8=35 \)

solving equations: Exercises 28,30,32,34,36,38,46,50 p. 214

\(y+4=7\) ...yields... \(y=3\)

\(2x=8 \) ...yields... \(x=4\)

\(4y+2=18 \) ...yields... \(y=4\)

\(1-6y=13 \) ...yields... \(y=-2\)

\(5-4s=6s-5 \) ...yields... \(s=1\)

\(3n-16=53 \) ...yields... \(n=23\)

\(p^{1/3}=3 \) ...yields... \(p=27\)

\(v^3+4=68 \) ...yields... \(v=4\)

You deposit \(\$1,200\) in an account with an annual interest rate of 3%. Calculate the amount of money you will have after 5 years, assuming that you earn simple interest.

Solution. Every year you earn \(0.03 \times 1200 = \$36 \). After 5 years you will earn \(5 \times 36=\$ 180 \). Your accumulated balance will be \(1200 + 180 = \$1380\).

Use the compound interest formula to compute the balance if \(\$ 10,000 \) is invested for 20 years in an account with an APR of 2.5% compounded annually

Solution.
The formula

\(A = P \times (1+APR)^Y \)
gives us

\(A=10000 \times (1+0.025)^{20} = \$ 16386.16 \)

What is the accumulated balance if \(\$2000 \) is invested for 5 years with an APR of 3% and daily compounding?

Solution.
The compound interest formula for interest paid \(n\) times a year is

\(A = P \times (1+\frac{APR}{n})^{(nY)} \)

In this case, n=365, and

\(A = 2000 \times (1+\frac{0.03}{365})^{(365 \times 5)} = \$ 2323.65\) <=== answer

To compare with continuous compounding

\(A = P \times e^{(APR \times Y)} \)
\(= 2000 \times e^{(0.03 \times 5)} = \$ 2323.67\)

Find the APY if a bank offers an APR of 3.2% compounded monthly

Solution.
Invest \(\$100 \), and after 1 year you will have

\(A = P \times (1+\frac{APR}{n})^{(nY)} = 100 \times (1+ \frac{0.032}{12})^{(12 \times 1)} = \$ 103.247\)

The relative difference is \( \frac{103.247-100}{100} =0.03247 =3.247 \%\)

Compute the balance after 1,5 and 20 years for a \(\$2000 \) deposit in an account with APR of 3.1%, and continuous compounding

Solution.
Continuous compounding formula

\(A = P\times e^{(APR \times Y)} \)

gives us

\(A = 2000\times e^{(0.031 \times Y)} \).

After 1 year: \(A = 2000\times e^{(0.031 \times 1)} = \$ 2062.97\)

After 5 years: \(A = 2000\times e^{(0.031 \times 5)} = \$ 2335.32\)

After 20 year: \(A = 2000\times e^{(0.031 \times 20)} = \$ 3717.86\)

How much must you deposit today into an account with quarterly compounding and an APR of 4.5% in order to have \(\$ 25,000 \) in 8 years?

Solution.
Compound interest formula for interest paid \(n\) times a year

\(A = P \times (1+\frac{APR}{n})^{(nY)} \)

solved for the principal

\(P= \frac{A}{(1+\frac{APR}{n})^{(nY)} }\)

gives us

\(P= \frac{25000}{(1+\frac{0.045}{4})^{(4\times 8)} } = \$17477\)

How much must you deposit today into an account with an APR of 5.5% compounded daily in order to have a \(\$120,000\) college fund in 15 years?

Solution.
Compound interest formula for interest paid \(n\) times a year

\(A = P \times (1+\frac{APR}{n})^{(nY)} \)

solved for the principal

\(P= \frac{A}{(1+\frac{APR}{n})^{(nY)} }\)

gives us

\(P= \frac{120000}{(1+\frac{0.055}{365})^{(365\times 15)} } = \$52,591.47\)

Jose invests \(\$1500\) into an account that earns \(5.6\%\) compounded annually. Marta invests \(\$1500\) into an account that earns \(5.7\%\) compounded annually. Compute the balance after \(10\) and \(30\) years.

Solution.
Compound interest formula for interest paid once a year is

\(A = P \times (1+APR)^Y \).

After \(10\) years:

Jose: \(1500 \times (1+0.056)^{10} = \$ 2,586.6 \)

Marta: \(1500 \times (1+0.057)^{10} = \$ 2,611.2 \)

Jose invests \(\$1500\) into an account that earns \(5.6\%\) compounded annually. Marta invests \(\$1500\) into an account that earns \(5.7\%\) compounded annually. Compute the balance after \(10\) and \(30\) years.

Solution.
Compound interest formula for interest paid once a year is

\(A = P \times (1+APR)^Y \).

After \(30\) years:

Jose: \(1500 \times (1+0.056)^{30} = \$ 7,691.5 \)

Marta: \(1500 \times (1+0.057)^{30} = \$ 7.913 \)

Jose invests \(\$1500\) into an account that earns \(5.6\%\) compounded annually. Marta invests \(\$1500\) into an account that earns \(5.7\%\) compounded annually. Compare the balance after \(10\) years.

Comparison.

After \(10\) years Jose has \(\$ 2,586.6 \), while Marta has \(\$ 2,611.2 \).

Relative difference \(=\frac{2,611.2 - 2,586.6}{2,586.6} \)
\(= \frac{24.6}{2,586.6} \)
\(\approx 0.0095 = 0.95\%\)

After \(10\) years, Marta has \(\$24.6\) or \(0.95\%\) more than Jose.

Jose invests \(\$1500\) into an account that earns \(5.6\%\) compounded annually. Marta invests \(\$1500\) into an account that earns \(5.7\%\) compounded annually. Compare the balance after \(30\) years.

Comparison.

After \(30\) years Jose has \(\$ 7,691.5 \), while Marta has \(\$ 7.913 \).

Relative difference \(=\frac{7913 - 7691.5}{7691.5} \)
\(= \frac{221.5}{7691.5} \)
\(\approx 0.029 = 2.9\%\)

After \(30\) years, Marta has \(\$ 221.5\) or \(2.9\%\) more than Jose.