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shannondidier

  • 2 years ago

Which of the following is the solution of log x + 50.001 = -3 ?

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  1. shannondidier
    • 2 years ago
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    here are the answer choices x = 5 x = 7 x = 13 x = 15

  2. swissgirl
    • 2 years ago
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    Well Lets try to isolate x So first we subtract 50.001 from both sides logx=-53.001

  3. shannondidier
    • 2 years ago
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    i typed it wrong, its base x+5 to the 0.001-3

  4. swissgirl
    • 2 years ago
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    Then we know that \( e^{ \log{x}} =x\) So \( e^{ \log{x}}=e^{-53.001}\)

  5. swissgirl
    • 2 years ago
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    hmmmm Can you retype it please cuz I dont think I am following?

  6. whpalmer4
    • 2 years ago
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    better yet, draw it!

  7. vinnv226
    • 2 years ago
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    @shannondidier What should we use as the base of the unspecified "log?" Some people interpret this as log base 10, others interpret it as the natural log (base e)

  8. shannondidier
    • 2 years ago
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    |dw:1373413752114:dw|

  9. swissgirl
    • 2 years ago
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    Seems like the base is x+5 not 10

  10. whpalmer4
    • 2 years ago
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    ah, that's much different :-)

  11. swissgirl
    • 2 years ago
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    yaaaaaaaaaaaa

  12. whpalmer4
    • 2 years ago
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    so it is asking us to find a value \(x\) such that \[(x+5)^{-3} = 0.001\]

  13. whpalmer4
    • 2 years ago
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    Here's a hint: \[x^{-n} = \frac{1}{x^n}\] and \[0.001 = \frac{1}{1000}\]

  14. shannondidier
    • 2 years ago
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    yeah i understand that. so whats the final answer?

  15. whpalmer4
    • 2 years ago
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    okay, look at your answer choices. do any of them + 5 = a number that when raised to the -3 power = 0.001, or when raised to the 3 power = 1000?

  16. whpalmer4
    • 2 years ago
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    maybe work backwards — what number cubed gives you 1000?

  17. shannondidier
    • 2 years ago
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    x = 5 x = 7 x = 13 x = 15 would it be x=5

  18. whpalmer4
    • 2 years ago
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    it would! \[(5+5)^{-3} = 10^{-3} = \frac{1}{10^3} = \frac{1}{1000} = 0.001\]

  19. shannondidier
    • 2 years ago
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    thank you!!!

  20. whpalmer4
    • 2 years ago
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    next time, please don't make us do a warm-up problem ;-)

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