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AMC 12/AHSME
1991 AMC 12/AHSME
20
Exponents
Exponents
Source: AHSME 1991 problem 20
October 30, 2011
geometry
algebra
polynomial
Vieta
logarithms
AMC
Problem Statement
The sum of all real
x
x
x
such that
(
2
x
−
4
)
3
+
(
4
x
−
2
)
3
=
(
4
x
+
2
x
−
6
)
3
(2^{x} - 4)^{3} + (4^{x} - 2)^{3} = (4^{x} + 2^{x} - 6)^{3}
(
2
x
−
4
)
3
+
(
4
x
−
2
)
3
=
(
4
x
+
2
x
−
6
)
3
is
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
3
/
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
5
/
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
7
/
2
<span class='latex-bold'>(A)</span>\ 3/2\qquad<span class='latex-bold'>(B)</span>\ 2\qquad<span class='latex-bold'>(C)</span>\ 5/2\qquad<span class='latex-bold'>(D)</span>\ 3\qquad<span class='latex-bold'>(E)</span>\ 7/2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
3/2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
5/2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
7/2
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