How to add exponents - SAT Math

Although each base is different, we can convert them to a common base of

We know

,

,

and

.

Remember to apply the power rule of exponents.

Therefore we have

.

We can factor out to get

.

When adding exponents, you want to factor out to make solving the question easier.

We can factor out to get

.

Now we can add exponents and therefore our answer is

.

Express as a power of ; that is: .

Then .

Using the properties of exponents, .

Therefore, , so .

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

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):

Step 1:

Step 2:

Step 3:

Step 1:

Step 2: The above is equal to

Similarly

So

Rewrite the term on the left as a product. Remember that negative exponents shift their position in a fraction (denominator to numerator).

The term on the right can be rewritten, as 27 is equal to 3 to the third power.

Exponent rules dictate that multiplying terms allows us to add their exponents, while one term raised to another allows us to multiply exponents.

We now know that the exponents must be equal, and can solve for .