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mar1e0721 Group Title

log 7 √ 49

  • one year ago
  • one year ago

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  1. satellite73 Group Title
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    first off, \(\sqrt{49}=7\) right?

    • one year ago
  2. mar1e0721 Group Title
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    \[\sqrt{7} 49\]

    • one year ago
  3. satellite73 Group Title
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    so you are being asked for \[\log_7(7)\] which must be 1, because you are being asked "raise 7 to what power to get 7?" and the answer is 1

    • one year ago
  4. satellite73 Group Title
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    now i am lost

    • one year ago
  5. satellite73 Group Title
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    what is the base of your log? what is the input?

    • one year ago
  6. amorfide Group Title
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    if you have this as base 10 then you are multiplying 7 by 7=49 log49 but if it is base 7, follow what @satellite73 said

    • one year ago
  7. mar1e0721 Group Title
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    log sqrt 7 bqase 49

    • one year ago
  8. mar1e0721 Group Title
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    Base**

    • one year ago
  9. amorfide Group Title
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    lol what?

    • one year ago
  10. satellite73 Group Title
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    \[\log_{49}(\sqrt{7})\]?

    • one year ago
  11. satellite73 Group Title
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    base is 49, input is \(\sqrt{7}\) right?

    • one year ago
  12. satellite73 Group Title
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    trick is to try to write with exponents, because \[\log_b(x)=y\iff b^y=x\] so you are trying to solve \[49^y=\sqrt{7}\]

    • one year ago
  13. satellite73 Group Title
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    we see that \(49=7^2\) and \(\sqrt{7}=7^{\frac{1}{2}}\) so what you want is \[7^{2y}=7^{\frac{1}{2}}\] or \[2y=\frac{1}{2}\] or \[y=\frac{1}{4}\]

    • one year ago
  14. satellite73 Group Title
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    meaning is simpler english that the number \(\sqrt{7}\) is the fourth root of 49

    • one year ago
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