Exponents - SAT Math
Card 0 of 2288
Which of the following is equal to 410 + 410 + 410 + 410 + 411?
Which of the following is equal to 410 + 410 + 410 + 410 + 411?
We can start by rewriting 411 as 4 * 410. This will allow us to collect the like terms 410 into a single term.
410 + 410 + 410 + 410 + 411
= 410 + 410 + 410 + 410 + 4 * 410
= 8 * 410
Because the answer choices are written with a base of 2, we need to rewrite 8 and 4 using bases of two. Remember that 8 = 23, and 4 = 22.
8 * 410
= (23)(22)10
We also need to use the property of exponents that (ab)c = abc. We can rewrite (22)10 as 22x10 = 220.
(23)(22)10
= (23)(220)
Finally, we must use the property of exponents that ab * ac = ab+c.
(23)(220) = 223
The answer is 223.
We can start by rewriting 411 as 4 * 410. This will allow us to collect the like terms 410 into a single term.
410 + 410 + 410 + 410 + 411
= 410 + 410 + 410 + 410 + 4 * 410
= 8 * 410
Because the answer choices are written with a base of 2, we need to rewrite 8 and 4 using bases of two. Remember that 8 = 23, and 4 = 22.
8 * 410
= (23)(22)10
We also need to use the property of exponents that (ab)c = abc. We can rewrite (22)10 as 22x10 = 220.
(23)(22)10
= (23)(220)
Finally, we must use the property of exponents that ab * ac = ab+c.
(23)(220) = 223
The answer is 223.
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If 3 + 3_n_+3 = 81, what is 3_n_+2 ?
If 3 + 3_n_+3 = 81, what is 3_n_+2 ?
3 + 3_n_+3 = 81
In this equation, there is a common factor of 3, which can be factored out.
Thus, 3(1 + 3_n_+2) = 81
Note: when 3 is factored out of 3_n_+3, the result is 3_n_+2 because (3_n_+3 = 31 * 3_n_+2). Remember that exponents are added when common bases are multiplied. Also remember that 3 = 31.
3(1 + 3_n_+2) = 81
(1 + 3_n_+2) = 27
3_n_+2 = 26
Note: do not solve for n individually. But rather seek to solve what the problem asks for, namely 3n+2.
3 + 3_n_+3 = 81
In this equation, there is a common factor of 3, which can be factored out.
Thus, 3(1 + 3_n_+2) = 81
Note: when 3 is factored out of 3_n_+3, the result is 3_n_+2 because (3_n_+3 = 31 * 3_n_+2). Remember that exponents are added when common bases are multiplied. Also remember that 3 = 31.
3(1 + 3_n_+2) = 81
(1 + 3_n_+2) = 27
3_n_+2 = 26
Note: do not solve for n individually. But rather seek to solve what the problem asks for, namely 3n+2.
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Simplify: (x3 * 2x4 * 5y + 4y2 + 3y2)/y
Simplify: (x3 * 2x4 * 5y + 4y2 + 3y2)/y
Let's do each of these separately:
x3 * 2x4 * 5y = 2 * 5 * x3 * x4 * y = 10 * x7 * y = 10x7y
4y2 + 3y2 = 7y2
Now, rewrite what we have so far:
(10x7y + 7y2)/y
There are several options for reducing this. Remember that when we divide, we can "distribute" the denominator through to each member. That means we can rewrite this as:
(10x7y)/y + (7y2)/y
Subtract the y exponents values in each term to get:
10x7 + 7y
Let's do each of these separately:
x3 * 2x4 * 5y = 2 * 5 * x3 * x4 * y = 10 * x7 * y = 10x7y
4y2 + 3y2 = 7y2
Now, rewrite what we have so far:
(10x7y + 7y2)/y
There are several options for reducing this. Remember that when we divide, we can "distribute" the denominator through to each member. That means we can rewrite this as:
(10x7y)/y + (7y2)/y
Subtract the y exponents values in each term to get:
10x7 + 7y
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If
, then which of the following is equivalent to
?
If , then which of the following is equivalent to
?
We can break up the equation into two smaller equations involving only x and y. Then, once we solve for x and y, we can find the value of
.

We can break up the equation into two smaller equations involving only x and y. Then, once we solve for x and y, we can find the value of .
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If
, what is the value of
?
If , what is the value of
?
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.
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.
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Simplify: y3x4(yx3 + y2x2 + y15 + x22)
Simplify: y3x4(yx3 + y2x2 + y15 + x22)
When you multiply exponents, you add the common bases:
y4 x7 + y5x6 + y18x4 + y3x26
When you multiply exponents, you add the common bases:
y4 x7 + y5x6 + y18x4 + y3x26
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Solve for x.
23 + 2x+1 = 72
Solve for x.
23 + 2x+1 = 72
The answer is 5.
8 + 2x+1 = 72
2x+1 = 64
2x+1 = 26
x + 1 = 6
x = 5
The answer is 5.
8 + 2x+1 = 72
2x+1 = 64
2x+1 = 26
x + 1 = 6
x = 5
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If
, what is the value of
?
If , what is the value of
?
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
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
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If _a_2 = 35 and _b_2 = 52 then _a_4 + _b_6 = ?
If _a_2 = 35 and _b_2 = 52 then _a_4 + _b_6 = ?
_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
_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
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If
, what is the value of
?
If , what is the value of
?
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 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.
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54 / 25 =
54 / 25 =
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.
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.
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What is the value of
such that
?
What is the value of such that
?
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 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 .
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Which of the following is eqivalent to 5_b_ – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) , where b is a constant?
Which of the following is eqivalent to 5_b_ – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) , where b is a constant?
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.
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.
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If
and
are positive integers, and
, then what is
in terms of
?
If and
are positive integers, and
, then what is
in terms of
?
is equal to
which is equal to
. If we compare this to the original equation we get 
is equal to
which is equal to
. If we compare this to the original equation we get
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Solve for
:

Solve for :
Combining the powers, we get
.
From here we can use logarithms, or simply guess and check to get
.
Combining the powers, we get .
From here we can use logarithms, or simply guess and check to get .
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If
, then what does
equal?
If, then what does
equal?
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Simplify. All exponents must be positive.

Simplify. All exponents must be positive.
Step 1: 
Step 2: 
Step 3: (Correct Answer): 
Step 1:
Step 2:
Step 3: (Correct Answer):
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Simplify. All exponents must be positive.

Simplify. All exponents must be positive.
Step 1: 
Step 2: 
Step 3:
Step 1:
Step 2:
Step 3:
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Answer must be with positive exponents only.
Answer must be with positive exponents only.
Step 1:
Step 2: The above is equal to 
Step 1:
Step 2: The above is equal to
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