How to add exponents - PSAT Math
Card 0 of 105
If 
, what is the value of 
?
If
Tap to see back →
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
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 = ?
Tap to see back →
_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
If 
, what is the value of 
?
If 
Tap to see back →
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 
For 
So we have 
Or
Divide this by 
Thus 
*Hint: If you are really unsure, you could have plugged in the numbers and found that the first choice worked in the equation.
Solve for x.
23 + 2x+1 = 72
Solve for x.
23 + 2x+1 = 72
Tap to see back →
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
What is the value of 
such that 
?
What is the value of
Tap to see back →
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
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?
Tap to see back →
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.
If r and s are positive integers, and 25left ( $5^{r}$ right $)=5^{s-2}$, then what is s in terms of r?
If r and s are positive integers, and 25left ( $5^{r}$ right $)=5^{s-2}$, then what is s in terms of r?
Tap to see back →
25left ( $5^{r}$ right ) is equal to 
which is equal to left ( $5^{r+2}$ right ). If we compare this to the original equation we get r+2=s-2rightarrow s=r+4
25left ( $5^{r}$ right ) is equal to
Solve for x:
Solve for x:
Tap to see back →
Combining the powers, we get $1024=2^{x}$.
From here we can use logarithms, or simply guess and check to get x=10.
Combining the powers, we get $1024=2^{x}$.
From here we can use logarithms, or simply guess and check to get x=10.
$Ifx^2$=11, then what does $x^4$ equal?
$Ifx^2$=11, then what does $x^4$ equal?
Tap to see back →
Simplify. All exponents must be positive.
left ( $x^{-2}$$y^{3}$ right )left ( $x^{5}$$y^{-4}$ right )
Simplify. All exponents must be positive.
left ( $x^{-2}$$y^{3}$ right )left ( $x^{5}$$y^{-4}$ right )
Tap to see back →
Step 1: left ( $x^{-2}$$x^{5}$ right )= $x^{3}$
Step 2: left ( $y^{3}$$y^{-4}$ right )= $y^{-1}$= $\frac{1}{y}$
Step 3: (Correct Answer): $$\frac{x^{3}$$}{y}
Step 1: left ( $x^{-2}$$x^{5}$ right )= $x^{3}$
Step 2: left ( $y^{3}$$y^{-4}$ right )= $y^{-1}$= $\frac{1}{y}$
Step 3: (Correct Answer): $$\frac{x^{3}$$}{y}
Simplify. All exponents must be positive.
Simplify. All exponents must be positive.
Tap to see back →
Step 1: $$\frac{y^{5}$$}{left ( $x^{3}$$x^{2}$ right )left right $)y^{-1}$}
Step 2: $\frac{left ( $y^{5}$$$y^{1}$ right $)}{x^{3}$$x^{2}$}
Step $3:$\frac{y^{6}$$$}{x^{5}$}
Step 1: $$\frac{y^{5}$$}{left ( $x^{3}$$x^{2}$ right )left right $)y^{-1}$}
Step 2: $\frac{left ( $y^{5}$$$y^{1}$ right $)}{x^{3}$$x^{2}$}
Step $3:$\frac{y^{6}$$$}{x^{5}$}
$\frac{left ( -11 right $)^{-8}$$}{left ( -11right $)^{12}$}
Answer must be with positive exponents only.
$\frac{left ( -11 right $)^{-8}$$}{left ( -11right $)^{12}$}
Answer must be with positive exponents only.
Tap to see back →
Step 1:$\frac{1}{left ( -11 right $)^{12}$$left ( -11 right $)^{8}$}
Step 2: The above is equal to $\frac{1}{left ( -11 right $)^{20}$$}
Step 1:$\frac{1}{left ( -11 right $)^{12}$$left ( -11 right $)^{8}$}
Step 2: The above is equal to $\frac{1}{left ( -11 right $)^{20}$$}
Evaluate:
-left ( -3 right $)^{0}$-left ( $-3^{0}$ right )
Evaluate:
-left ( -3 right $)^{0}$-left ( $-3^{0}$ right )
Tap to see back →
-left ( -3 right $)^{0}$= -1
-left ( -3 right $)^{0}$= -1
Simplify:
Simplify:
Tap to see back →
Similarly 
So 
Similarly
So
How many of the following base ten numbers have a base five representation of exactly four digits?
(A) 
(B) 
(C) 
(D) 
How many of the following base ten numbers have a base five representation of exactly four digits?
(A)
(B)
(C)
(D)
Tap to see back →
A number in base five has powers of five as its place values; starting at the right, they are 
The lowest base five number with four digits would be
in base ten.
The lowest base five number with five digits would be
in base ten.
Therefore, a number that is expressed as a four-digit number in base five must fall in the range
Three of the four numbers - all except 100 - fall in this range.
A number in base five has powers of five as its place values; starting at the right, they are
The lowest base five number with four digits would be
The lowest base five number with five digits would be
Therefore, a number that is expressed as a four-digit number in base five must fall in the range
Three of the four numbers - all except 100 - fall in this range.
If , what is the value of ?
If
Tap to see back →
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
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 = ?
Tap to see back →
_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
If , what is the value of ?
If
Tap to see back →
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
For
So we have
Or
Divide this by
Thus
*Hint: If you are really unsure, you could have plugged in the numbers and found that the first choice worked in the equation.
Solve for x.
23 + 2x+1 = 72
Solve for x.
23 + 2x+1 = 72
Tap to see back →
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
What is the value of such that ?
What is the value of
Tap to see back →
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