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shannondidier

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

  • 9 months ago
  • 9 months ago

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  1. shannondidier
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    here are the answer choices x = 5 x = 7 x = 13 x = 15

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

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

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

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

    • 9 months ago
  6. whpalmer4
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    better yet, draw it!

    • 9 months ago
  7. vinnv226
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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)

    • 9 months ago
  8. shannondidier
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    |dw:1373413752114:dw|

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

    • 9 months ago
  10. whpalmer4
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    ah, that's much different :-)

    • 9 months ago
  11. swissgirl
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    yaaaaaaaaaaaa

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

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

    • 9 months ago
  14. shannondidier
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    yeah i understand that. so whats the final answer?

    • 9 months ago
  15. whpalmer4
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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?

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

    • 9 months ago
  17. shannondidier
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    x = 5 x = 7 x = 13 x = 15 would it be x=5

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

    • 9 months ago
  19. shannondidier
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    thank you!!!

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

    • 9 months ago
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