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Melissa_Something

  • one year ago

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  1. Melissa_Something
    • one year ago
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    \[\log _{10} 1/\sqrt{10}\]

  2. Melissa_Something
    • one year ago
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    Sqrt 10/ 10 is wrong :(

  3. Nnesha
    • one year ago
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    \[\log_{10} \frac{1}{\sqrt{10}}\] like this ?

  4. Jhannybean
    • one year ago
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    Change of base: \(\log_a (x) =\dfrac{\log_b (x)}{\log_b (a)}\)

  5. Melissa_Something
    • one year ago
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    Oh yes! Lol

  6. Nnesha
    • one year ago
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    you can convert square root to an exponent if you want or use change of base formula

  7. jango_IN_DTOWN
    • one year ago
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    log(a/b)=log a-log b so the given expression becomes log 1-log sqrt 10 =-log sqrt 10 =-log (10)^1/2=-1/2

  8. Jhannybean
    • one year ago
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    Lets take \(x=\dfrac{1}{\sqrt{10}}\)

  9. Nnesha
    • one year ago
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    \[\sqrt{y} \] can be written as \[\rm y^\frac{ 1 }{ 2 }\] so \[\rm \log_{10} \frac{ 1 }{ (10)^\frac{ 1 }{ 2 } }\] move the (10)^/2 at the numerator

  10. Melissa_Something
    • one year ago
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    So it would be \[\log 1/\sqrt{10} /\log 10\] ??

  11. Melissa_Something
    • one year ago
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    Oh oh okay

  12. Jhannybean
    • one year ago
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    \[\begin{align} \log_{10} \left(\frac{1}{\sqrt{10}}\right) &=\frac{\log\left(\frac{1}{\sqrt{10}}\right)}{\log(10)}\\& = \frac{\log(1)-\log(\sqrt{10})}{\log(10)} \\&=\frac{-\frac{1}{2}\log(10)}{\log(10)}\\&=-\frac{1}{2} \end{align}\]

  13. Melissa_Something
    • one year ago
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    Wow thanks for all the support :o

  14. Nnesha
    • one year ago
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    typo (10)^{1/2} remember the exponent rule when move base from the numerator to denominator sign of the exponent would change \[\huge\rm \frac{ 1 }{ x^{-m }}= x^m\]

  15. Melissa_Something
    • one year ago
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    Yes oh my gosh thank you @Nnesha

  16. Nnesha
    • one year ago
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    u already got the answer so i'm just gonna work itout \[\log_{10} 10^{-\frac{1}{2}}\] apply the power rule power rule \[\large\rm log_b x^y = y \log_b x\] \[-\frac{1}{2} \log_{10}10\] log_{10} 10 = 1 so left wth -1/2

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