CONTOH SOAL EKSPONEN DAN LOGARITMA

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PEMBAHASAN:

1. Jawab: C

( 1 + 3 ) – ( 4 – ) = ( 1 + 3 ) – ( 4 – ) = 1 + 3 – 4 + 5

= – 3 + 8

2. Jawab: B

15_{log 20 =}

=

3. Jawab: A

= - -

-= ( –5 . –1 . –3 ) = –15 ( 1 )

= –15

4. Jawab: B

–

– – –

=

–

– – –

=

–

(5)

5

=

–

=

–

x

=

= 9 ( 2 + 1 )

5. Jawab: E

32x+1– 28 . 3x + 9 = 0 32x . 3 – 28 . 3x + 9 = 0 ( misal : 3x = A ) 3A2 – 28A + 9 = 0

( 3A – 1 ) ( A – 9 ) = 0 A =

A = 9

A = 3x = A = 3x = 9

3x = 3–1 atau 3x = 32

x2 = –1 x1 = 2

Sehingga: 3x1 – x2 = 3 ( 2 ) – ( –1 )= 7

6. Jawab: B

2 . 34x – 20 . 32x + 18 = 0 misal: 32x = A

2A2 _–_{20A + 18 = 0 ( dibagi dengan 2 )} A2– 10A + 9 = 0

( A – 1 ) ( A – 9 ) = 0 A = 1 A = 9

A = 32x = 1 A = 32x = 9 32x = 30 atau 32x = 32

x = 0 x = 1

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7. Jawab: A

2 . 9x– 3x+1 + 1 = 0

2 . 32x – 3x . 3 + 1 = 0 ( misal: 3x = A ) 2A2 – 3A + 1 = 0

( 2A – 1 ) ( A – 1 ) = 0 A =

A = 1

8. Jawab: A

2

log [ 2log ( 2x+1 + 3 ) ] = 1 + 2log x = 2log 2 + 2log x

2

log [ 2log ( 2x+1 + 3 ) ] = 2log 2x

2log ( 2x+1 + 3 ) = 2x

2x+1 + 3 = ( 2 )2x

22x– 2x . 2 – 3 = 0 ( misal: 2x = A ) A2 – 2A – 3 = 0

( A – 3 ) ( A + 1 ) = 0 A = 3 A = –1

A = 2x = 3 atau A = 2x = –1 x = 2log 3 x = { }

9. Jawab: C

log f ( x ) < log g ( x ) f ( x ) < g ( x ) ; f ( x ) > 0 ; g ( x ) > 0

log ( x – 4 ) + log ( x + 8 ) < log ( 2x + 16 ) log ( x – 4 ) ( x + 8 ) < log ( 2x + 16 ) log ( x2 + 4x – 32 ) < log ( 2x + 16 ) x2 + 4x – 32 < 2x + 16

x2 + 2x – 48 < 0 ( x + 8 ) ( x – 6 ) < 0

Mengecek nilai f ( x ) > 0: Mengecek nilai g ( x ) > 0:

x2 + 4x – 32 > 0 2x + 16 > 0 ( x + 8 ) ( x – 4 ) > 0 2x > –16

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Himpunan penyelesaian dari soal ini harus irisan semua himpunan:

10. Jawab: C

log f ( x ) < log g ( x ) f ( x ) < g ( x ) ; f ( x ) > 0 ; g ( x ) > 0

2 . log x log ( 2x + 5 ) + 2 . log 2 log x2 log ( 2x + 5 ) + log 22

log x2 log 4 ( 2x + 5 )

log x2 log ( 8x + 20 )

x2 8x + 20

x2– 8x – 20 0 ( x – 10 ) ( x + 2 ) 0

Mengecek nilai f ( x ): Mengecek nilai g ( x ):

x2 > 0 8x + 20 > 0

x > 0 8x > –20

x <

11. Jawab: E

> _– – > _–

– >

–2x > 36 x < –18

Sehingga himpunan penyelesaian:

HP = { x | 4 < x < 6 }

Sehingga himpunan penyelesaian:

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8

12. Jawab: B

log f ( x ) = log g ( x ) f ( x ) = g ( x ) ; f ( x ) > 0 ; g ( x ) > 0 ; a > 0

x_{log ( 10x}3_–_{9x ) =}x_{log x}5 10x3 – 9x = x5 x5 – 10x3 + 9x = 0 x ( x4 _–_10x2_{+ 9 ) = 0} x ( x2 – 9 ) ( x2– 1 ) = 0 x ( x + 3 ) ( x – 3 ) ( x + 1 ) ( x – 1 ) = 0 x = 0 x = –3 x = –1 x = 1 x = 3

Mengecek syarat f ( x ) > 0: Mengecek syarat g ( x ) > 0 dan a:

10x3– 9x > 0 g ( x ): x5 > 0 x > 0

x ( 10x2 – 9 ) > 0 a: x > 0

x = 0 x =

Sehingga himpunan penyelesainnya adalah { 1, 3 }

13. Jawab: B

– – – –

– –

x2– 3x + 4 < 2x – 2 x2– 5x + 6 < 0 ( x – 2 ) ( x – 3 ) < 0

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9

14. Jawab: E

(3_{log x)}2_–_{3 .}3_{log x + 2 = 0 ( misal:}3_{log x = A )} A2– 3A + 2 = 0

( A – 2 ) ( A – 1 ) = 0 A = 2 A = 1

A = 3_{log x = 2} _{A =}3_{log x = 1} x = 32 _atau _{x = 3}1 x = 9 x = 3

Sehingga x1 . x2 = ( 9 ) ( 3 ) = 27

15. Jawab: E

–

> – – > –

–2 + x > – –12 + 6x > 5x – 5 x > 7

16. Jawab: A

a

log f ( x ) < alog g ( x ) f ( x ) < g ( x ) ; f ( x ) < 0 ; g ( x ) < 0

2

log ( x2 – 3x + 2 ) < 2log ( 10 – x ) x2– 3x + 2 < 10 – x x2– 2x – 8 < 0 ( x – 4 ) ( x + 2 ) < 0

Mengecek f ( x ) > 0: Mengecek g ( x ): x2 _–_{3x + 2 > 0} ₁₀_–_{x > 0} ( x – 1 ) ( x – 2 ) > 0 – x > –10

x > 10

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17. Jawab: B

a_{log f ( x ) < b} _{f ( x ) < a}b_{; f ( x ) > 0}

9

log ( x2 + 2x ) < x2 + 2x <

x2 + 2x < 3

x2 + 2x – 3 < 0 ( x + 3 ) ( x – 1 ) < 0

Mengecek syarat f ( x ) > 0: Himpunan penyelesaian: x2 + 2x > 0

x ( x + 2 ) > 0

18. Jawab: A

Diketahui: 2x + 2–x = 5

22x + 2–2x = ( 2x )2 + ( 2–x )2

= ( 2x + 2–x )2– 2 ( 2x . 2–x ) = ( 5 )2– 2 ( 1 )

= 25 – 2 = 23

19. Jawab: D

2x + 4 =

6x + 12 = 4x + 20

2x = 8

x = 4

Sehingga 2x = 24 = 16

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20. Jawab: E

log ( x – 1 )2_{< log ( x}_–_{1 )} _{( x}_–_{1 )}2_{< ( x}_–_{1 )} x2– 2x + 1 < x – 1 x2– 3x + 2 < 0 ( x – 1 ) ( x – 2 ) < 0

Mengecek syarat f ( x ) > 0: Mengecek syarat g ( x ) > 0:

x2 – 2x + 1 > 0 x – 1 > 0

( x – 1 ) ( x – 1 ) > 0 x > 1

CONTOH SOAL EKSPONEN DAN LOGARITMA

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