Respuesta :
Answer:
To solve the equation
2
�
−
2
+
3
3
−
�
=
3
2
x−2
+3
3−x
=3, we'll first try to rewrite it in a more convenient form. Notice that
3
=
3
1
3=3
1
, so we can rewrite the equation as:
2
�
−
2
+
3
3
−
�
=
3
1
2
x−2
+3
3−x
=3
1
Now, we can apply the property of exponents that states
�
�
+
�
=
�
�
⋅
�
�
a
b+c
=a
b
⋅a
c
. So, we can rewrite the equation as:
2
�
−
2
+
3
3
3
�
=
3
1
2
x−2
+
3
x
3
3
=3
1
2
�
−
2
+
27
3
�
=
3
2
x−2
+
3
x
27
=3
Now, let's express
2
�
−
2
2
x−2
as
2
�
2
2
2
2
2
x
:
2
�
4
+
27
3
�
=
3
4
2
x
+
3
x
27
=3
Multiplying both sides by 4 to eliminate the fraction, we get:
2
�
+
108
3
�
=
12
2
x
+
3
x
108
=12
Multiplying both sides by
3
�
3
x
to eliminate the second fraction, we get:
2
�
⋅
3
�
+
108
=
12
⋅
3
�
2
x
⋅3
x
+108=12⋅3
x
2
�
⋅
3
�
−
12
⋅
3
�
+
108
=
0
2
x
⋅3
x
−12⋅3
x
+108=0
Now, let
�
=
3
�
y=3
x
, then the equation becomes a quadratic equation:
2
�
2
−
12
�
+
108
=
0
2y
2
−12y+108=0
Divide the equation by 2:
�
2
−
6
�
+
54
=
0
y
2
−6y+54=0
Now, solve this quadratic equation for
�
y. Using the quadratic formula:
�
=
−
�
±
�
2
−
4
�
�
2
�
y=
2a
−b±
b
2
−4ac
where
�
=
1
a=1,
�
=
−
6
b=−6, and
�
=
54
c=54:
�
=
6
±
(
−
6
)
2
−
4
⋅
1
⋅
54
2
⋅
1
y=
2⋅1
6±
(−6)
2
−4⋅1⋅54
�
=
6
±
36
−
216
2
y=
2
6±
36−216
�
=
6
±
−
180
2
y=
2
6±
−180
Since the discriminant is negative, there are no real solutions for
�
y, which means there are no real solutions for
3
�
3
x
, and consequently, no real solutions for
�
x. Thus, the equation
2
�
−
2
+
3
3
−
�
=
3
2
x−2
+3
3−x
=3 has
Step-by-step explanation:
Now, solve this quadratic equation for
�
y. Using the quadratic formula:
�
=
−
�
±
�
2
−
4
�
�
2
�
y=
2a
−b±
b
2
−4ac
where
�
=
1
a=1,
�
=
−
6
b=−6, and
�
=
54
c=54:
�
=
6
±
(
−
6
)
2
−
4
⋅
1
⋅
54
2
⋅
1
y=
2⋅1
6±
(−6)
2
−4⋅1⋅54
�
=
6
±
36
−
216
2
y=
2
6±
36−216
�
=
6
±
−
180
2
y=
2
6±
−180
