Resources
Testimonials
Plans
Sign in
Sign up
Resources
Testimonials
Plans
Home
Math Problems
Precalculus
Find the roots of factored polynomials
Solve for
x
x
x
.
\newline
Enter the solutions from least to greatest.
\newline
(
2
x
−
1
)
(
x
+
4
)
=
0
(2x-1)(x+4)=0
(
2
x
−
1
)
(
x
+
4
)
=
0
\newline
lesser
\newline
x
=
x=
x
=
\newline
greater
\newline
x
=
x=
x
=
Get tutor help
Simplify
e
ln
3
+
3
e^{\ln 3+3}
e
l
n
3
+
3
and write without any logarithms.
\newline
Answer:
Get tutor help
Simplify
e
ln
2
+
2
e^{\ln 2+2}
e
l
n
2
+
2
and write without any logarithms.
\newline
Answer:
Get tutor help
Simplify
e
ln
2
−
ln
4
e^{\ln 2-\ln 4}
e
l
n
2
−
l
n
4
and write without any logarithms.
\newline
Answer:
Get tutor help
Find the value of
x
x
x
that solves the equation
ln
(
x
−
5
)
−
ln
2
=
0
\ln (x-5)-\ln 2=0
ln
(
x
−
5
)
−
ln
2
=
0
.
\newline
Answer:
Get tutor help
Find the value of
x
x
x
that solves the equation
ln
(
x
−
3
)
−
ln
3
=
0
\ln (x-3)-\ln 3=0
ln
(
x
−
3
)
−
ln
3
=
0
.
\newline
Answer:
Get tutor help
Find the value of
x
x
x
that solves the equation
ln
(
x
−
2
)
−
2
ln
4
=
ln
1
\ln (x-2)-2 \ln 4=\ln 1
ln
(
x
−
2
)
−
2
ln
4
=
ln
1
.
\newline
Answer:
x
=
x=
x
=
Get tutor help
Simplify
e
ln
4
−
ln
2
e^{\ln 4-\ln 2}
e
l
n
4
−
l
n
2
and write without any logarithms.
\newline
Answer:
Get tutor help
Find the value of
x
x
x
that solves the equation
ln
(
x
+
4
)
−
1
3
ln
27
=
ln
7
\ln (x+4)-\frac{1}{3} \ln 27=\ln 7
ln
(
x
+
4
)
−
3
1
ln
27
=
ln
7
.
\newline
Answer:
x
=
x=
x
=
Get tutor help
Find the value of
x
x
x
that solves the equation
ln
(
x
+
2
)
−
1
3
ln
8
=
ln
2
\ln (x+2)-\frac{1}{3} \ln 8=\ln 2
ln
(
x
+
2
)
−
3
1
ln
8
=
ln
2
.
\newline
Answer:
x
=
x=
x
=
Get tutor help
Find the value of
x
x
x
that solves the equation
ln
(
x
−
1
)
+
1
2
ln
16
=
ln
2
\ln (x-1)+\frac{1}{2} \ln 16=\ln 2
ln
(
x
−
1
)
+
2
1
ln
16
=
ln
2
.
\newline
Answer:
x
=
x=
x
=
Get tutor help
Let
f
(
x
)
=
x
2
−
2
x
−
1
f(x)=\frac{x^{2}-2}{x-1}
f
(
x
)
=
x
−
1
x
2
−
2
.
\newline
f
′
(
x
)
=
f^{\prime}(x)=
f
′
(
x
)
=
Get tutor help
Let
g
(
x
)
=
x
−
5
x
2
+
1
g(x)=\frac{x-5}{x^{2}+1}
g
(
x
)
=
x
2
+
1
x
−
5
.
\newline
g
′
(
x
)
=
g^{\prime}(x)=
g
′
(
x
)
=
Get tutor help
Let
h
(
x
)
=
x
2
+
1
2
x
2
−
3
x
h(x)=\frac{x^{2}+1}{2 x^{2}-3 x}
h
(
x
)
=
2
x
2
−
3
x
x
2
+
1
.
\newline
h
′
(
x
)
=
h^{\prime}(x)=
h
′
(
x
)
=
Get tutor help
Let
f
(
x
)
=
x
−
4
f(x)=x-4
f
(
x
)
=
x
−
4
and
g
(
x
)
=
(
x
−
4
)
3
.
g(x)=(x-4)^{3} \text {. }
g
(
x
)
=
(
x
−
4
)
3
.
\newline
Find the sum of the areas enclosed by the graphs of
f
f
f
and
g
g
g
between
x
=
3
x=3
x
=
3
and
x
=
5
x=5
x
=
5
.
\newline
Use a graphing calculator and round your answer to three decimal places.
Get tutor help
Previous
1
...
5
6
