Resources
Testimonials
Plans
Sign in
Sign up
Resources
Testimonials
Plans
AI tutor
Welcome to Bytelearn!
Let’s check out your problem:
8
5
÷
8
3
)
×
8
−
2
\left.8^{5} \div 8^{3}\right) \times 8^{-2}
8
5
÷
8
3
)
×
8
−
2
View step-by-step help
Home
Math Problems
Algebra 1
Evaluate integers raised to rational exponents
Full solution
Q.
8
5
÷
8
3
)
×
8
−
2
\left.8^{5} \div 8^{3}\right) \times 8^{-2}
8
5
÷
8
3
)
×
8
−
2
Apply Quotient Rule:
Use the quotient rule for exponents:
a
m
/
a
n
=
a
(
m
−
n
)
a^{m}/a^{n} = a^{(m-n)}
a
m
/
a
n
=
a
(
m
−
n
)
.
\newline
Calculate
8
5
÷
8
3
8^{5}\div8^{3}
8
5
÷
8
3
.
\newline
8
(
5
−
3
)
=
8
2
8^{(5-3)} = 8^2
8
(
5
−
3
)
=
8
2
Calculate
8
5
÷
8
3
8^{5}\div8^{3}
8
5
÷
8
3
:
Now multiply
8
2
8^2
8
2
by
8
−
2
8^{-2}
8
−
2
using the product rule for exponents:
a
m
×
a
n
=
a
m
+
n
a^{m} \times a^{n} = a^{m+n}
a
m
×
a
n
=
a
m
+
n
.
\newline
Calculate
8
2
×
8
−
2
8^2 \times 8^{-2}
8
2
×
8
−
2
.
\newline
8
2
+
(
−
2
)
=
8
0
8^{2 + (-2)} = 8^0
8
2
+
(
−
2
)
=
8
0
Apply Product Rule:
Recognize that any non-zero number raised to the power of
0
0
0
is
1
1
1
.
\newline
Calculate
8
0
8^0
8
0
.
\newline
8
0
=
1
8^0 = 1
8
0
=
1
More problems from Evaluate integers raised to rational exponents
Question
Simplify.
\newline
8
1
1
2
81^{\frac{1}{2}}
8
1
2
1
\newline
_____________
Get tutor help
Posted 2 months ago
Question
Simplify.
\newline
(
−
32
)
1
/
5
(-32)^{1/5}
(
−
32
)
1/5
\newline
______
Get tutor help
Posted 2 months ago
Question
Simplify. Assume all variables are positive.
\newline
w
4
3
w
7
3
\frac{w^{\frac{4}{3}}}{w^{\frac{7}{3}}}
w
3
7
w
3
4
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
Get tutor help
Posted 2 months ago
Question
Simplify. Assume all variables are positive.
\newline
(
2
r
1
3
)
8
(2r^{\frac{1}{3}})^8
(
2
r
3
1
)
8
\newline
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
Get tutor help
Posted 1 month ago
Question
Simplify. Assume all variables are positive.
\newline
v
7
4
⋅
v
9
4
v^{\frac{7}{4}} \cdot v^{\frac{9}{4}}
v
4
7
⋅
v
4
9
\newline
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
\newline
Get tutor help
Posted 2 months ago
Question
Simplify.
\newline
1
1
4
⋅
1
1
4
1^{\frac{1}{4}} \cdot 1^{\frac{1}{4}}
1
4
1
⋅
1
4
1
\newline
______
Get tutor help
Posted 1 month ago
Question
Simplify.
\newline
3
2
(
−
1
/
5
)
32^{(-1/5)}
3
2
(
−
1/5
)
Get tutor help
Posted 1 month ago
Question
Simplify. Assume all variables are positive.
\newline
w
3
2
w
5
2
\frac{w^{\frac{3}{2}}}{w^{\frac{5}{2}}}
w
2
5
w
2
3
\newline
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
\newline
Get tutor help
Posted 2 months ago
Question
Simplify. Assume all variables are positive.
\newline
(
27
x
)
2
3
(27x)^{\frac{2}{3}}
(
27
x
)
3
2
\newline
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
Get tutor help
Posted 2 months ago
Question
Simplify. Assume all variables are positive.
\newline
x
1
4
x
11
4
\frac{x^{\frac{1}{4}}}{x^{\frac{11}{4}}}
x
4
11
x
4
1
\newline
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
\newline
Get tutor help
Posted 1 month ago