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 $\displaystyle -3+6=3$.