Exponents - SAT Math

(xy)4 can be rewritten as x4y4 and z0 = 1 because a number to the zero power equals 1. After simplifying, you get 1/y.

When dividing expressions with the same variable, combine terms by subtracting the exponents, while leaving the variable unchanged. For this problem, we do that by subtracting 6-2, to get a new exponent of 4:

Multiplying two exponents that have the same base is the equivalent of simply adding the exponents.

So is the same as , and if , then or .

Properties of exponents suggests that when multiplying the same base, add the exponents, when dividing, subtract the exponents on bottom from those on top, and when raising an exponent to another power, multiply the exponents. Remember that (x4/x5) = x–1 = 1/x; Still using order of operations (PEMDAS) we get the following:(x4y2/x5)3= (y2/x)3 = y6/(x3).

x7 / x-3/2 = x7 (x+3/2) based on the fact that division changes the sign of an exponent.

x7 (x+3/2) = x7+3/2 due to the additive property of exponent numbers that are multiplied.

7+3/2= 14/2 + 3/2 = 17/2 so

x7 / x-3/2 = x7+3/2 = x17/2

Since x7 / x-3/2 = xn, xn = x17/2

So n = 17/2

Using exponents, 27 is equal to 33. So, the equation can be rewritten:

34_x_ + 6 = (33)2_x_

34_x_ + 6 = 36_x_

When both side of an equation have the same base, the exponents must be equal. Thus:

4_x_ + 6 = 6_x_

6 = 2_x_

x = 3

_a_4 = _a_2 * _a_2 and _b_6= _b_2 * _b_2 * _b_2

Therefore _a_4 + _b_6 = 35 * 35 + 52 * 52 * 52 = 1,225 + 140,608 = 141,833

1. According to the rules of exponents, one can add the exponents when adding to variables with the same base. So, x2x4 becomes x6.

2. The rules of exponents also state that if the bases are the same, one can substract the exponents when dividing. So, x6/x becomes x5. Similarly, y/y2 becomes 1/y.

3. When combining these operations, one gets x5/y.

25 = 5 * 5 = 52. Then 54 / 25 = 54 / 52.

Now we can subtract the exponents because the operation is division. 54 / 52 = 54 – 2 = 52 = 25. The answer is therefore 25.

Since we have two ’s in we will need to combine the two terms.

For this can be rewritten as

So we have .

Or

Divide this by :

Thus or

*Hint: If you are really unsure, you could have plugged in the numbers and found that the first choice worked in the equation.

Since the base is 5 for each term, we can say 2 + n =12. Solve the equation for n by subtracting 2 from both sides to get n = 10.

When you multiply exponents, you add the common bases:

y4 x7 + y5x6 + y18x4 + y3x26

The answer is 5.

8 + 2x+1 = 72

2x+1 = 64

2x+1 = 26

x + 1 = 6

x = 5

We can solve by converting all terms to a base of two. 4, 16, and 32 can all be expressed in terms of 2 to a standard exponent value.

We can rewrite the original equation in these terms.

Simplify exponents.

Finally, combine terms.

From this equation, we can see that .

We want to simplify 5_b_ – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) .

Notice that we can collect the –5(b–1) terms, because they are like terms. There are 5 of them, so that means we can write –5(b–1) – 5(b–1) – 5(b–1) – 5(b–1) – 5(b–1) as (–5(b–1))5.

To summarize thus far:

5_b_ – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–1) = 5_b +(–5(_b–_1))5

It's important to interpret –5(b–1) as (–1)5(b–1) because the –1 is not raised to the (b – 1) power along with the five. This means we can rewrite the expression as follows:

5_b_ +(–5(b–1))5 = 5_b_ + (–1)(5(b–1))(5) = 5_b_ – (5(b–1))(5)

Notice that 5(b–1) and 5 both have a base of 5. This means we can apply the property of exponents which states that, in general, abac = a b+c. We can rewrite 5 as 51 and then apply this rule.

5_b_ – (5(_b–1))(5) = 5_b – (5(_b–1))(51) = 5_b – 5(_b–_1+1)

Now, we will simplify the exponent b – 1 + 1 and write it as simply b.

5_b_ – 5(b–1+1) = 5_b – 5_b = 0

The answer is 0.

is equal to which is equal to . If we compare this to the original equation we get

Combining the powers, we get .

From here we can use logarithms, or simply guess and check to get .

Step 1:

Step 2:

Step 3: (Correct Answer):