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shannondidier

  • one year ago

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

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

  2. swissgirl
    • one year 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
    • one year ago
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    i typed it wrong, its base x+5 to the 0.001-3

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

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

  6. whpalmer4
    • one year ago
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    better yet, draw it!

  7. vinnv226
    • one year 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
    • one year ago
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    |dw:1373413752114:dw|

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

  10. whpalmer4
    • one year ago
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    ah, that's much different :-)

  11. swissgirl
    • one year ago
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    yaaaaaaaaaaaa

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

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

  14. shannondidier
    • one year ago
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    yeah i understand that. so whats the final answer?

  15. whpalmer4
    • one year 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
    • one year ago
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    maybe work backwards — what number cubed gives you 1000?

  17. shannondidier
    • one year ago
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    x = 5 x = 7 x = 13 x = 15 would it be x=5

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

  19. shannondidier
    • one year ago
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    thank you!!!

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

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