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Penyelesaian pert1daksamaan 2 lo g ( x + 1 ) ≤ lo g ( x + 4 ) + 4 adalah ...

Penyelesaian pert1daksamaan adalah ...

  1. x space less or equal than space 7

  2. x > -4

  3. -1 < x < 5

  4. negative 1 space less or equal than space x space less than space 6

  5. x greater or equal than 6

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2 space log left parenthesis x space plus space 1 right parenthesis space less or equal than space log space space left parenthesis x space plus space 4 right parenthesis space plus space 4  space log left parenthesis x space plus space 1 right parenthesis squared space less or equal than space log space space left parenthesis x space plus space 4 right parenthesis space plus space 4  left parenthesis x space plus space 1 right parenthesis squared space less or equal than space left parenthesis x space plus space 4 right parenthesis space plus space 4  x squared space plus space 2 x space plus space 1 space less or equal than space 4 x space plus space 16  x squared space minus space 2 x space minus space 15 space less or equal than space 0  space left parenthesis x space minus space 5 right parenthesis left parenthesis x space plus space 3 right parenthesis space less or equal than space 0 space  x space equals space 5 space a d a l a h space x space equals space minus 3    S y a r a t space n u m e r u s colon  space x plus 1 greater than 1  space x space greater than space minus 1 space  x plus 4 greater than 0  space x space greater than space minus 4 space  M a k a space h i m p u n a n space p e n y e. l e s a i n n y a space a d a l a h space x space greater than space minus 4

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