Primer Control Acumulativo Cuarto Mediosolucion
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SOLUCIÓN PRIMER CONTROL ACUMULATIVO NOMBRE: __________________________________________ CURSO: ____________ FECHA: ___________ I.
Reduce las siguientes potencias:
1.
[( 3x)
2.
( 27
3.
2
( )
⋅ 5x 3
p −1
] = [9x ⋅ 25x ]
2 3
⋅ 93− p
) = [(3 ) 2
[ y ⋅ (3 y ) ] : 9 y 2 2 2
2
6 3
2
3 p −1
4
[
]
3
= 225 x 8 = 2253 x 24
⋅ (32 ) 3− p
[
] = [3 2
]
3 p −3
⋅ 36 − 2 p
] = [3 ] 2
[ ]
2
p +3 2
= 32 p + 6
2
= y 2 ⋅ 9 y 4 : 9 y 4 = 9 y 6 : 9 y 4 = 81y12 : 9 y 4 = 9 y 8
II. Resuelve las siguientes ecuaciones exponenciales: 4.
Si 9 ⋅ 9 = 3 x , entonces x = ∴ 81 = 3 X ⇒ 34 = 3 X ⇒ 4 = x
x 2
x 2
−1
2x
1
5.
2 1 2 ⋅ 2 = , entonces x = ? ∴ 2 2 = ⇒ 2 x = 21 ⇒ x = 1 2 1
6.
3 x +1 ⋅ 3 x −1 =
1 1 1 1 2x −1 , entonces x + 1 = ∴ 3 = 3 ⇒ 2 x = −1 ⇒ x = − ⇒ x + 1 = − + 1 = 3 2 2 2
Reduce las siguientes potencias:
1.
[( 3x)
2.
( 27
3.
2
( )
⋅ 5x 3
p −1
] = [9x ⋅ 25x ]
2 3
⋅ 93− p
) = [(3 ) 2
[ y ⋅ (3 y ) ] : 9 y 2 2 2
2
6 3
2
3 p −1
4
[
]
3
= 225 x 8 = 2253 x 24
⋅ (32 ) 3− p
[
] = [3 2
]
3 p −3
⋅ 36 − 2 p
] = [3 ] 2
[ ]
2
p +3 2
= 32 p + 6
2
= y 2 ⋅ 9 y 4 : 9 y 4 = 9 y 6 : 9 y 4 = 81y12 : 9 y 4 = 9 y 8
II. Resuelve las siguientes ecuaciones exponenciales: 4.
Si 9 ⋅ 9 = 3 x , entonces x = ∴ 81 = 3 X ⇒ 34 = 3 X ⇒ 4 = x
x 2
x 2
−1
2x
1
5.
2 1 2 ⋅ 2 = , entonces x = ? ∴ 2 2 = ⇒ 2 x = 21 ⇒ x = 1 2 1
6.
3 x +1 ⋅ 3 x −1 =
1 1 1 1 2x −1 , entonces x + 1 = ∴ 3 = 3 ⇒ 2 x = −1 ⇒ x = − ⇒ x + 1 = − + 1 = 3 2 2 2
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