Each year, you can calculate your interest by multiplying the principle (\(\displaystyle \$5000\)) by \(\displaystyle 1.025\). For one year, this would be:

\(\displaystyle 1.025*5000=5125\)

For two years, it would be:

\(\displaystyle 5125*1.025\), which is the same as \(\displaystyle 1.025*1.025*5000\)

Therefore, you can solve for a five year period by doing:

\(\displaystyle 1.025^5*5000\)

Using your calculator, you can expand the \(\displaystyle 1.025^5\) into a series of multiplications. This gives you \(\displaystyle 5657.041064453125\), which is closest to \(\displaystyle \$5657\). 

It is easiest to break this down into steps. For each year, you will multiply by \(\displaystyle 1.035\) to calculate the new value. Therefore, let's make a chart:

After year 1: \(\displaystyle 7500*1.035=7762.5\); Total interest: \(\displaystyle 262.5\)

After year 2: \(\displaystyle 7762.5*1.035=8034.1875\); Let us round this to \(\displaystyle 8034.19\); Total interest: \(\displaystyle 271.69\)

After year 3: \(\displaystyle 8034.19 * 1.035 = 8315.38665\); Let us round this to \(\displaystyle 8315.39\); Total interest: \(\displaystyle 281.2\)

Thus, the positive difference of the interest from the last period and the interest from the first period is: \(\displaystyle 281.2-271.69=9.51\)

First, break the problem into two segments: the amount Jack invests in the high-yield savings, and the amount Jack invests in the simple interest account (10,000 and 5,000 respectively).

Now let's work with the high-yield savings account. $10,000 is invested at an annual rate of 8%, compounded quarterly. We can use the compound interest formula to solve:

\(\displaystyle \text{Final Balance}=\text{ Principal} \cdot \bigg(1+\frac{\text{Interest Rate}}{c}\bigg)^{(\text{Time})(c)}\)

Plug in the values given:

\(\displaystyle =10,000\cdot \bigg(1+\frac{.08}{4}\bigg)^{(1)(4)}\)

\(\displaystyle =10,000 \cdot (1.02)^{4}\)

\(\displaystyle =10,824.32\)

Therefore, Jack makes $824.32 off his high-yield savings account. Now let's calculate the other interest:

\(\displaystyle \text{Interest}=\text{Principal} \cdot \text{ Interest Rate} \cdot \text{ Time}\)

\(\displaystyle =5,000 \cdot (0.06)\cdot (1)\)

\(\displaystyle =300\)

Add the two together, and we see that Jack makes a total of, \(\displaystyle \$1124.32\) off of his investments.

The first step is to convert the depreciation rate into a decimal. \(\displaystyle 13\%=0.13\). Now lets recall the exponential decay model. \(\displaystyle y=C(1-r)^t\), where \(\displaystyle C\) is the starting amount of money, \(\displaystyle r\) is the annual rate of decay, and \(\displaystyle t\) is time (in years). After substituting, we get

\(\displaystyle y=20000(1-0.13)^8\)

\(\displaystyle y=20000(0.87)^8\)

\(\displaystyle y=20000\cdot 0.3282117\)

\(\displaystyle y= 6564.23\)

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 \(\displaystyle -3+6=3\).