Laws Of Exponents Flashcards
a small number that is written on the upper right of another number (or variable) which is called the base
Exponent
Example 1:Compute 35
Solution:The exponent of 5 tells us that 3 is multiplied by itself five times.
Therefore, 35= 243
Example 1:What is the value of -94?
Solution:-94is an example of Case 1. To compute for -94, we need to calculate 94first. Afterward, multiply the result by -1:
6561
Example 2:What is the value of (- 9)4?
Solution:Since – 9 is inside the parentheses, it indicates that – 9 is what is being raised to the power of 4. Thus, we need to multiply – 9 to itself four times.
6561
Example 1:Write k5in expanded form.
Solution:The exponent in k5tells us that the variable k is being multiplied by itself 5 times. Thus, the expanded form of k5is:
k5= k∙k∙k∙k∙k
Example 2:Expressu∙u∙u∙u∙u∙u in exponential form.
Solution:Note that the variableuis used six times. Hence, we must use an exponent of 6. Thus:
u∙u∙u∙u∙u∙u= u6
Example 1:Write -3x5in expanded form.
Solution:The variablexis the only one raised to the power of 5. Thus, only the variablexis the base of exponent 5 in the given, and -3 is not included.
Thus, -3x5= -3(x∙x∙x∙x∙x)
Example 2:Write (-3x)5in expanded form.
Solution:The existence of the parentheses indicates that both the -3 and x in -3xare raised to the power of 5. Thus, -3x is the base of exponent 5.
In other words, (-3x)5= -3x∙–3x∙ -3x∙–3x∙ -3x
Suppose we want to multiplyx2byx4. Note thatx2andx4have the same base.How can we multiply them?
One possible method is to expandx2andx4:
x2= x∙x
x4= x∙x∙x∙x
Multiplying the expanded values:
x2∙x4
(x∙x)∙( x∙x∙x∙x)
Note that we can express the product of the expanded values into exponential form:
(x∙x)∙( x∙x∙x∙x) =x∙x∙x∙x∙x∙x =x6
Therefore,x2∙x4=x6
Example 1:Compute for 2⁴∙2²
Solution:We have expressions with the same bases (i.e., 2) being multiplied together. Thus, we can apply the product rule.
Let us copy the common base first:
Then, add the exponents:
Thus, using the product rule:2⁴∙2²= 2⁶
Example 2:Multiply b⁵by b³
Solution:Since we have the same bases being multiplied together, we can apply the product rule:
Let us copy the common base first:
Then, add the exponents:
Therefore, using the product rule:b⁵∙b³=b⁸
Example 3:Multiply a³b²by a²b⁴
Solution:We have two bases involved here, the variablesaandb.
Thus, we need to apply the product rule, each foraandb:
Let us copy the common bases first:
Add the exponents for the common bases.
Hence,a³b²∙a²b⁴=a⁵b⁶
Example 4:Multiply (x + 5)⁶by (x + 5)³
Solution:In this case, the common base isx + 5.Hence, we can apply the product rule.
Let us copy the common base first:
Add the exponents:
Therefore,(x + 5)⁶∙(x + 5)³= (x + 5)⁹
Example 5:Compute for a(a²)
Solution: If two variables are written together with the other one enclosed in parentheses, it implies that the variables are being multiplied. Since we have a common base in the given (which isa), we can apply the product rule here.
Note that if a number or a variable has no exponent written above it, it implies that the exponent is 1.
Let us copy the common base first:
Add the exponents:
Therefore,a(a²) = a³
Example 1:Compute for x⁷÷x³
Solution:Since we are dividing exponential expressions with the same base, we can apply the quotient rule.
Let us copy the common base first:
Subtract the exponents:
Therefore,x⁷÷x³= x⁴
Example 2:Simplify x⁹⁄x⁴
Solution:x9⁄x4also means x9÷ x4. Since we are dividing exponential expressions with the same base, we can apply the quotient rule.
Let us start by copying the common base:
Finally, subtract the exponents:
Therefore, x⁹/x⁴= x⁵
Example 3:Simplify p⁸q²⁄p⁶q
Solution:We have two bases involved: the variablespandq. Thus, we will use the quotient rule for the variablespandq.
Copying the common bases:
Finally, subtract the exponents for each of the common bases.
Hence,p8q2⁄p6q= p²q
Example 4:Divide 1 000 000 000 by 1 000 000
Solution:Note that we can express1 000 000 000 as 109. On the other hand, we can express 1 000 000 as 106. Therefore, we can answer the problem by dividing 109by 106.
Since we have a common base (which is 10), we can apply the quotient rule:
Let us copy the common base first:
Then, subtract the exponents:
Therefore, the answer is 103or 1000.
Tip:We can express a multiple of 10 into exponential form quickly by counting the number of zeros it has. For example, 1 000 000 000 has 9 zeros. Thus, if we express 1 000 000 000 in exponential form, we can determine the exponent to be used based on the number of zeros it has. Therefore 1 000 000 000 = 10⁹
Example 1:Simplify (k⁴)²
Solution:Notice that the entirek4is raised to 2. Applying the power rule, we can combine the exponents into one by multiplying them. Thus,
(k4)2= k4×2= k8
Therefore,(k4)2= k⁸
Example 2:What is the value of (32)3?
Solution:Applying the product rule:
(32)3= 32×3= 36
Now, all we need to do next is expand 36:
3 ∙ 3 ∙ 3 ∙ 3 ∙ 3 ∙ 3 = 729
Therefore,(32)3= 729
Example 3:Simplify (a⁵)²
Solution:Applying the product rule:
(a5)2= a5×2= a10
Therefore, (a5)2= a¹⁰
Example 4:Simplify (8y⁵)²
Solution:Note that the base of the exponent 2 is 8y5. This means that we need to apply the power rule both for 8 and y5:
(8y5)2= 82y5×2= 82y10
Since 82= 8 ∙ 8 = 64, then 82y10= 64y10
Therefore, (8y5)2= 64y¹⁰
If an expression has more than one variable multiplied together and raised to a certain power, we can simplify that expression using the
Power of a product rule
Example 1:Simplify (xy)²
Solution:The given expression has two variables multiplied together (which isxy)and raised to the power of 2. This means that we can simplify it using the power of a product rule.
The power of the product rule allows us to “distribute” the exponent to each variable:
(xy)2 = (x2)(y2)
Therefore,(xy)2= x²y²
Example 2:Simplify (a⁴b³)³
Solution:Let us apply the power of the product rule to simplify the given expression.
We start by “distributing” 3 to each of the variables:
(a4b3)3= (a4)3(b3)3
To further simplify the expression, we can apply the product rule to each variable:
(a4)3(b3)3= a4×3b3×3= a12b9
Hence,(a4b3)3= a¹²b⁹
Example 3:Simplify (4a³b²)²
Solution:Let us apply the power of the product rule to simplify the given expression. Note that we should also raise 4 to the power of 2:
(4a3b2)2= (4)2(a3)2(b2)2
Now, we apply the product rule to each variable:
(4)2(a6)(b4)
Lastly, since 42= 16:
16a6b4
Therefore, (4a3b2)2= 16a⁶b⁴
if variables are divided together and raised to a certain power, we can simplify the expression using the
Power of the quotient rule
Example 1:Apply the power of the quotient rule to (x⁄y)²
Solution:Through the power of the quotient rule, we can distribute the exponent to the variables involved in the division process:
Therefore, using the power of the quotient rule, (x⁄y)2= x²⁄y²
Example 2:Simplify (a⁴⁄b²)²
Solution:Since we have two variables (a4andb2) divided together and raised to a certain power, we can apply the power of the quotient rule:
Notice that we can simplify the expression further using the product rule:
Therefore, (a4⁄b2)2= a⁸⁄b⁴
The Zero-Exponent Rule states that any nonzero base raised to
0
Example 1:Suppose that m≠0, what is the value of m⁰?
Solution:By the zero-exponent rule,m⁰= 1.
Example 2:What is the value of 109⁰?
Solution:By the zero-exponent rule, 109⁰= 1
Example 3:Simplify 15x⁰
Solution:We know that by the zero-exponent rule,x0= 1. Take note thatx0is multiplied by 15 in 15x0. Sincex0= 1:
Hence, 15x⁰= 15
Example 4:Simplify a⁰b²c
Solution:By the zero-exponent rule,a0= 1. Sincea0is multiplied tob2cin the given expressiona0b2c:
Hence,a0b2c = b²c
Example 1:What is the value of 5^-3?
Solution:Using the negative exponent rule, we can express 5-3as 1⁄53. Note that we can expand 53as 5 x 5 x 5 and obtain 125. Therefore, 5^-3= 1⁄125
Example 2:Express y^-1as an expression without a negative exponent.
Solution:Using the negative exponent rule, we can express y-1as 1⁄y which has no negative exponent involved.
Thus, the answer is 1⁄y.
Example 3:Express a^2b^-3c without a negative exponent.
Solution:We can apply the negative exponent rule to express a2b-3c without a negative exponent. However, since b– 3is the only base raised to a negative exponent, then we can only apply the negative exponent rule to b-3and it is the only base that we are going to put in the denominator.
Therefore, a2b-3ccan be written without a negative exponent as a^2c⁄b^3
a^m∙a^n= a^m + n
Product rule
a^m⁄a^n= a^m – n
Quotient rule
(a^m)^n= a^mn
Power rule
(a b)^p= a^pb^p
Power of a Product Rule
(a⁄b)^m= a^m⁄b^m, whereb≠ 0
Power of a Quotient Rule
a^0= 1
Zero Exponent Rule
a^-m= 1⁄a^m
Negative Exponent Rule
Example 1:Simplify 5p^0q^-2
Solution:We can simplify the given expression by making all of its exponents positive.
Let us start by applying the zero-exponent rule:
5p0q-2
5(1)q-2(since p0= 1)
5q-2
We can then remove the negative exponent using the negative exponent Rule:
5q-2
5⁄q2
Thus, the answer is5⁄q^2
Example 2:Simplify (a^4⁄a^2)^2
Solution 1:Note that we can distribute the exponent 2 which is outside the parentheses to the bases that are inside the parentheses using the power of the quotient rule:
(a4⁄a2)2= a4×2⁄a2×2= a8⁄a4
Since we are dividing the same bases, we can apply the quotient rule:
a8⁄a4 =a8 – 4= a4
Therefore,(a4⁄a2)2= a4
You can also simplify the given expression using the alternative solution below.
Solution 2:This time, let us start applying the quotient rule since we are dividing the same bases:
(a4⁄a2)2= (a4 – 2)2= (a2)2
Notice that we can apply now the power rule since (a2)2is an expression raised to an exponent then raised to another exponent.
(a2)2= a2×2= a^4
Example 3:Simplify [(x + y)^2(x + y)^3]^-1
Solution:We can start by making the negative exponent positive. To do this, put the base into the denominator (negative exponent rule). The base in the given expression is the entire(x + y)2(x + y)3
We can simplify the expression further using the product rule since we are multiplying the same bases:
Therefore, the answer is 1⁄(x + y)^5
1) Compute 3a^2b^3•ab^2
a) 3a^3b^5
b) 3ab^5
c) 3a^2b^5
d) 3ab
3) Express a^-1b^-3c^2 as an expression without negative exponent
a)𝑎𝑐^2/𝑏^3
b)𝑎𝑏^3/𝑐^2
c) ab^3c^2
d)𝑐^2/𝑎𝑏^3
4) Express (k + m)^2(k + m)^-1 as an expression without any negative exponent
a) (k + m)^2
b) (k + m)^-1
c) k + m
d) k^2 + m^2
5) Apply the Laws of Exponents to compute for the value of (3 • 10^2)(3 • 10^3)
a) 900 000
b) 9 000 000
c) 90 000
d) 9 000
