The formula for compound interest is

\(\displaystyle A = A_{0}\left ( 1+ \frac{r}{N} \right )^{Nt}\)

where \(\displaystyle A_{0}\) is the initial investment, \(\displaystyle r\) is the interest rate expressed as the decimal equivalent, \(\displaystyle N\) is the number of periods per year the interest is compounded, \(\displaystyle t\) is the number of years, and \(\displaystyle A\) is the final value of the investment.

Set \(\displaystyle A_{0} = 10,000, r = 0.08125, N = 12\) (monthly = 12 periods), and \(\displaystyle t = 5\), and evaluate \(\displaystyle A\):

\(\displaystyle A = 10,000 \left ( 1+ \frac{0.08125}{12} \right )^{12 \cdot 5}\)

\(\displaystyle A \approx 10,000 \left ( 1.0067708\right )^{60}\)

\(\displaystyle A \approx 10,000 \cdot 1.499124\)

\(\displaystyle A \approx 1 4,991.24\) 

The CD will be worth $14,991.24.

The formula for compound interest is

\(\displaystyle A = A_{0}\left ( 1+ \frac{r}{N} \right )^{Nt}\)

where \(\displaystyle A_{0}\) is the initial investment, \(\displaystyle r\) is the interest rate expressed as the decimal equivalent, \(\displaystyle N\) is the number of periods per year the interest is compounded, \(\displaystyle t\) is the number of years, and \(\displaystyle A\) is the final value of the investment.

In the given scenario, \(\displaystyle t = 10\), \(\displaystyle r = 0.06735\), and \(\displaystyle N = 12\) (monthly); substitute:

\(\displaystyle A = A_{0}\left ( 1+ \frac{0.06735}{12} \right )^{12 \cdot 10}\)

\(\displaystyle A \approx A_{0}\left ( 1+0.0056125 \right )^{12 0}\)

\(\displaystyle A \approx A_{0}\left ( 1 .0056125 \right )^{12 0}\)

\(\displaystyle A \approx A_{0} \cdot 1.9574\)

This meas that the final value of the bonds is about 1.96 times their initial value, or, equivalently, 96% greater than their initial value. Of the given responses, 95% comes closest.

The formula for compound interest is

\(\displaystyle A = A_{0}\left ( 1+ \frac{r}{N} \right )^{Nt}\)

where \(\displaystyle A_{0}\) is the initial investment, \(\displaystyle r\) is the interest rate expressed as the decimal equivalent, \(\displaystyle N\) is the number of periods per year the interest is compounded, \(\displaystyle t\) is the number of years, and \(\displaystyle A\) is the final value of the investment.

Set \(\displaystyle A = 100,000, r = 0.07885, N = 12\) (monthly = 12 periods), and \(\displaystyle t = 10\), and evaluate \(\displaystyle A_{0}\):

\(\displaystyle 100,000 = A_{0} \left ( 1+ \frac{0.07885}{12} \right )^{12 \cdot 10}\)

\(\displaystyle 100,000 \approx A_{0} \left ( 1+0.0065708\right )^{120}\)

\(\displaystyle 100,000 \approx A_{0} \left ( 1 .0065708\right )^{120}\)

\(\displaystyle 100,000 \approx A_{0} \cdot 2.1944\)

\(\displaystyle A_{0} \approx 100,000 \div 2.1944\)

\(\displaystyle A_{0} \approx 45,569.99\)

The correct response is $45,569.99.

In order to find the interest earned, used the compound interest formula

\(\displaystyle \text{Final Balance} = \text{Principal} \cdot (1+\frac{IR}{C})^{(\text{Time})(\text{C})}\)

where \(\displaystyle c\) represents the number of times the account is compounded each year, and \(\displaystyle IR\) represents the interest rate expressed as a decimal.

\(\displaystyle =15,000\cdot(1+\frac{.06}{2})^{(2)(2)}\)

\(\displaystyle =15,000\cdot(1+.03)^{4}\)

\(\displaystyle =15,000\cdot(1.03)^{4}\)

\(\displaystyle =16882.63\)

The account is worth $16882.63 after two years. Therefore Tom earns $1882.63 in interest.

\(\displaystyle 16882.63-15000=1882.63\)

Multiplying like bases means add the exponents, so x+4 = 12, or x = 8.

Raising a power to a power means multiply the exponents, so 3y = 15, or y = 5.

x - y = 8 - 5 = 3.

The first step is to express both sides of the equation with equal bases, in this case 3. The equation becomes 3^3p = 3^2q. So then 3p = 2q, and q = (3/2)p is our answer. 

27^2/3 is 27 squared and cube-rooted. We want to pick the easier operation first. Here that is the cube root. To see that, try both operations. 

27^2/3 = (27²)^1/3 = 729^1/3 OR

27^2/3 = (27^1/3)² = 3²

Obviously 3² is much easier. Either 3² or 729^1/3 will give us the correct answer of 9, but with 3² it is readily apparent. 

To solve this problem, we will have to take the log of both sides to bring down our exponents. By doing this, we will get \(\displaystyle \dpi{100} \small a\ast log\left (\frac{1}{3} \right )= b\ast log\left ( 27 \right )\).

To solve for \(\displaystyle \dpi{100} \small \frac{a}{b}\) we will have to divide both sides of our equation by \(\displaystyle \dpi{100} \small log\frac{1}{3}\) to get \(\displaystyle \dpi{100} \small \frac{a}{b}=\frac{log\left ( 27 \right )}{log\left ( \frac{1}{3} \right )}\).

\(\displaystyle \dpi{100} \small \frac{log\left ( 27 \right )}{log\left ( \frac{1}{3} \right )}\) will give you the answer of –3.

We use two properties of logarithms: 

\(\displaystyle log(xy) = log (x) + log (y)\)

\(\displaystyle log(x^{n}) = nlog (x)\)

So \(\displaystyle \log 12=2 \log2+\log3\)

\(\displaystyle x^{m}\ast x^{n} = x^{m + n}\), here \(\displaystyle m=-3\) and \(\displaystyle n=6\), hence \(\displaystyle -3+6=3\).