Pattern Behaviors in Exponents - SAT Math

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:

Plug in the values given:

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

Add the two together, and we see that Jack makes a total of, off of his investments.

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

For two years, it would be:

, which is the same as

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

Using your calculator, you can expand the into a series of multiplications. This gives you , which is closest to .

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

After year 1: ; Total interest:

After year 2: ; Let us round this to ; Total interest:

After year 3: ; Let us round this to ; Total interest:

Thus, the positive difference of the interest from the last period and the interest from the first period is:

The first step is to convert the depreciation rate into a decimal. . Now lets recall the exponential decay model. , where is the starting amount of money, is the annual rate of decay, and is time (in years). After substituting, we get

The powers of i form a sequence that repeats every four terms.

i1 = i

i2 = -1

i3 = -i

i4 = 1

i5 = i

Thus:

i25 = i

i23 = -i

i21 = i

i19= -i

Now we can evalulate the expression.

i25 - i23 + i21 - i19 + i17..... + i

= i + (-1)(-i) + i + (-1)(i) ..... + i

= i + i + i + i + ..... + i

Each term reduces to +i. Since there are 13 terms in the expression, the final result is 13i.

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.

This problem is quite simple if you recall that the units place of powers of 2 follows a simple 4-step sequence.

Observe the first few powers of 2:

21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64, 27 = 128, 28 = 256 . . .

The units place follows a sequence of 2, 4, 8, 6, 2, 4, 8, 6, etc. Thus, divide 102 by 4. This gives a remainder of 2.

The second number in the sequence is 4, so the answer is 4.

We know that the expression must be negative. Therefore one or all of the terms x7, y8 and z10 must be negative; however, even powers always produce positive numbers, so y8 and z10 will both be positive. Odd powers can produce both negative and positive numbers, depending on whether the base term is negative or positive. In this case, x7 must be negative, so x must be negative. Thus, the answer is x < 0.

The first step is to express both sides of the equation with equal bases, in this case 3. The equation becomes 33p = 32q. So then 3p = 2q, and q = (3/2)p is our answer.

272/3 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.

272/3 = (272)1/3 = 7291/3 OR

272/3 = (271/3)2 = 32

Obviously 32 is much easier. Either 32 or 7291/3 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 .

To solve for we will have to divide both sides of our equation by to get .

will give you the answer of –3.

We use two properties of logarithms:

So

, here and , hence .

=

which means

Remember the laws of exponents. In particular, when the base is nonzero:

An effective way to compare these statements, is to convert them all into exponents with base 2. The original statement becomes:

This is identical to statement I. Now consider statement II:

Therefore, statement II is not identical to the original statement. Finally, consider statement III:

which is also identical to the original statement. As a result, only I and III are the same as the original statement.

Using properties of radicals e.g.,

we get

Properties of Radicals

Properties of Radicals

Properties of Radicals