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GoldRush18

Solve the equation below:

  • one year ago
  • one year ago

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  1. GoldRush18
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    \[\log_{4} x=1+\log_{2} 2x, x>0\]

    • one year ago
  2. GoldRush18
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    i guess we must solve for x

    • one year ago
  3. GoldRush18
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    @experimentX do u understand?

    • one year ago
  4. experimentX
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    yeah ... solve for x.

    • one year ago
  5. experimentX
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    use this property of log \[ \log_a b = \frac{\log_kb}{\log_k a}\]

    • one year ago
  6. GoldRush18
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    i still dont get it sorry

    • one year ago
  7. experimentX
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    sorry ... error there \[ \log_4x = \frac{\log_2k}{\log_24} = \frac{\log_2k}{2} \]

    • one year ago
  8. experimentX
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    Oops ... error again, that k is supposed to be x

    • one year ago
  9. experimentX
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    if you know other few properties of log then it's should be easy for you!!

    • one year ago
  10. experimentX
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    \[ \frac{\log_2x}{2} = \log_22 + \log_22x\] \[ \implies \log_2x = 2\log_22 + 2\log_22x \] \[ \implies \log_2x = \log_22^2 + \log_2(2x)^2 \] \[ \implies \log_2x = \log_2(2^2*(2x)^2) \] \[ \implies x = 2^2*(2x)^2 \]

    • one year ago
  11. experimentX
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    \[ \implies x = \frac{1}{16} \]

    • one year ago
  12. GoldRush18
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    @experimentX thanks

    • one year ago
  13. experimentX
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    yw ... though not sure about answer!! check on wolframalpha!!

    • one year ago
  14. GoldRush18
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    its correct

    • one year ago
  15. GoldRush18
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    because these are past papers they tell us the answer but they dont show us how to get it

    • one year ago
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