Logarithm rules i.e. $-1/2 cdot log_2(2/9)$ to $2/9cdot log_2(2)$?












1












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I'm struggling to understand the flow of calculation as shown in the picture below.



enter image description here



It would be really nice if someone can explain how does one reach from step one to step two and which Logarithm rules were applied here to reach the second step from the first one.



Thank you! :)






algebra-precalculus logarithms







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    $begingroup$
    While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
    $endgroup$
    – Matteo
    Jan 7 at 20:46








  • 3




    $begingroup$
    The substitution isn’t correct.
    $endgroup$
    – KM101
    Jan 7 at 20:48
















1












$begingroup$


I'm struggling to understand the flow of calculation as shown in the picture below.



enter image description here



It would be really nice if someone can explain how does one reach from step one to step two and which Logarithm rules were applied here to reach the second step from the first one.



Thank you! :)






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
    $endgroup$
    – Matteo
    Jan 7 at 20:46








  • 3




    $begingroup$
    The substitution isn’t correct.
    $endgroup$
    – KM101
    Jan 7 at 20:48














1












1








1





$begingroup$


I'm struggling to understand the flow of calculation as shown in the picture below.



enter image description here



It would be really nice if someone can explain how does one reach from step one to step two and which Logarithm rules were applied here to reach the second step from the first one.



Thank you! :)






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$




I'm struggling to understand the flow of calculation as shown in the picture below.



enter image description here



It would be really nice if someone can explain how does one reach from step one to step two and which Logarithm rules were applied here to reach the second step from the first one.



Thank you! :)






algebra-precalculus logarithms




algebra-precalculus logarithms






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share|cite|improve this question













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edited Jan 7 at 21:21









callculus

17.9k31427




17.9k31427










asked Jan 7 at 20:28









Giga2001Giga2001

103




103








  • 2




    $begingroup$
    While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
    $endgroup$
    – Matteo
    Jan 7 at 20:46








  • 3




    $begingroup$
    The substitution isn’t correct.
    $endgroup$
    – KM101
    Jan 7 at 20:48














  • 2




    $begingroup$
    While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
    $endgroup$
    – Matteo
    Jan 7 at 20:46








  • 3




    $begingroup$
    The substitution isn’t correct.
    $endgroup$
    – KM101
    Jan 7 at 20:48








2




2




$begingroup$
While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
$endgroup$
– Matteo
Jan 7 at 20:46






$begingroup$
While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
$endgroup$
– Matteo
Jan 7 at 20:46






3




3




$begingroup$
The substitution isn’t correct.
$endgroup$
– KM101
Jan 7 at 20:48




$begingroup$
The substitution isn’t correct.
$endgroup$
– KM101
Jan 7 at 20:48










1 Answer
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$begingroup$

I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.



begin{align}
&1 - frac{-frac{1}{3}log_2left(frac{1}{3}right) - frac{1}{2}log_2left(frac{2}{9}right)}{-sumlimits_{s = 1}^9 frac{1}{9}log_2left(frac{1}{9}right)}\\
&textrm{Denominator: You're just adding the same thing 9 times}\\
= &1 - frac{frac{1}{3}log_2left(left(frac{1}{3}right)^{-1}right) + frac{1}{2}log_2left(left(frac{2}{9}right)^{-1}right)}{-9 cdot frac{1}{9}log_2left(frac{1}{9}right)}\\
= &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{-log_2left(frac{1}{9}right)}\\
= &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(left(frac{1}{9}right)^{-1}right)}\\
= &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(9right)}\\
= &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}left(log_2left(9right) - log_2left(2right)right)}{log_2left(9right)}\\
= &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(9right) - frac{1}{2}}{log_2left(9right)}\\
end{align}



For this to be equal to the second step, you'd need $frac{1}{2}log_2left(9right) - frac{1}{2} = frac{2}{9}log_2left(2right) = frac{2}{9}$ to be true. It's not: $frac{1}{2}log_2left(9right) - frac{1}{2} approx 1.1$ and $frac{2}{9} approx 0.2$.






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    1 Answer
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    0












    $begingroup$

    I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.



    begin{align}
    &1 - frac{-frac{1}{3}log_2left(frac{1}{3}right) - frac{1}{2}log_2left(frac{2}{9}right)}{-sumlimits_{s = 1}^9 frac{1}{9}log_2left(frac{1}{9}right)}\\
    &textrm{Denominator: You're just adding the same thing 9 times}\\
    = &1 - frac{frac{1}{3}log_2left(left(frac{1}{3}right)^{-1}right) + frac{1}{2}log_2left(left(frac{2}{9}right)^{-1}right)}{-9 cdot frac{1}{9}log_2left(frac{1}{9}right)}\\
    = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{-log_2left(frac{1}{9}right)}\\
    = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(left(frac{1}{9}right)^{-1}right)}\\
    = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(9right)}\\
    = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}left(log_2left(9right) - log_2left(2right)right)}{log_2left(9right)}\\
    = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(9right) - frac{1}{2}}{log_2left(9right)}\\
    end{align}



    For this to be equal to the second step, you'd need $frac{1}{2}log_2left(9right) - frac{1}{2} = frac{2}{9}log_2left(2right) = frac{2}{9}$ to be true. It's not: $frac{1}{2}log_2left(9right) - frac{1}{2} approx 1.1$ and $frac{2}{9} approx 0.2$.






    share|cite|improve this answer









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      0












      $begingroup$

      I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.



      begin{align}
      &1 - frac{-frac{1}{3}log_2left(frac{1}{3}right) - frac{1}{2}log_2left(frac{2}{9}right)}{-sumlimits_{s = 1}^9 frac{1}{9}log_2left(frac{1}{9}right)}\\
      &textrm{Denominator: You're just adding the same thing 9 times}\\
      = &1 - frac{frac{1}{3}log_2left(left(frac{1}{3}right)^{-1}right) + frac{1}{2}log_2left(left(frac{2}{9}right)^{-1}right)}{-9 cdot frac{1}{9}log_2left(frac{1}{9}right)}\\
      = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{-log_2left(frac{1}{9}right)}\\
      = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(left(frac{1}{9}right)^{-1}right)}\\
      = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(9right)}\\
      = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}left(log_2left(9right) - log_2left(2right)right)}{log_2left(9right)}\\
      = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(9right) - frac{1}{2}}{log_2left(9right)}\\
      end{align}



      For this to be equal to the second step, you'd need $frac{1}{2}log_2left(9right) - frac{1}{2} = frac{2}{9}log_2left(2right) = frac{2}{9}$ to be true. It's not: $frac{1}{2}log_2left(9right) - frac{1}{2} approx 1.1$ and $frac{2}{9} approx 0.2$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.



        begin{align}
        &1 - frac{-frac{1}{3}log_2left(frac{1}{3}right) - frac{1}{2}log_2left(frac{2}{9}right)}{-sumlimits_{s = 1}^9 frac{1}{9}log_2left(frac{1}{9}right)}\\
        &textrm{Denominator: You're just adding the same thing 9 times}\\
        = &1 - frac{frac{1}{3}log_2left(left(frac{1}{3}right)^{-1}right) + frac{1}{2}log_2left(left(frac{2}{9}right)^{-1}right)}{-9 cdot frac{1}{9}log_2left(frac{1}{9}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{-log_2left(frac{1}{9}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(left(frac{1}{9}right)^{-1}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(9right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}left(log_2left(9right) - log_2left(2right)right)}{log_2left(9right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(9right) - frac{1}{2}}{log_2left(9right)}\\
        end{align}



        For this to be equal to the second step, you'd need $frac{1}{2}log_2left(9right) - frac{1}{2} = frac{2}{9}log_2left(2right) = frac{2}{9}$ to be true. It's not: $frac{1}{2}log_2left(9right) - frac{1}{2} approx 1.1$ and $frac{2}{9} approx 0.2$.






        share|cite|improve this answer









        $endgroup$



        I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.



        begin{align}
        &1 - frac{-frac{1}{3}log_2left(frac{1}{3}right) - frac{1}{2}log_2left(frac{2}{9}right)}{-sumlimits_{s = 1}^9 frac{1}{9}log_2left(frac{1}{9}right)}\\
        &textrm{Denominator: You're just adding the same thing 9 times}\\
        = &1 - frac{frac{1}{3}log_2left(left(frac{1}{3}right)^{-1}right) + frac{1}{2}log_2left(left(frac{2}{9}right)^{-1}right)}{-9 cdot frac{1}{9}log_2left(frac{1}{9}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{-log_2left(frac{1}{9}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(left(frac{1}{9}right)^{-1}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(9right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}left(log_2left(9right) - log_2left(2right)right)}{log_2left(9right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(9right) - frac{1}{2}}{log_2left(9right)}\\
        end{align}



        For this to be equal to the second step, you'd need $frac{1}{2}log_2left(9right) - frac{1}{2} = frac{2}{9}log_2left(2right) = frac{2}{9}$ to be true. It's not: $frac{1}{2}log_2left(9right) - frac{1}{2} approx 1.1$ and $frac{2}{9} approx 0.2$.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Jan 12 at 3:46









        PiKindOfGuyPiKindOfGuy

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