Simplification with exponents












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I'm currently revising an exam about channel coding in telecommunications and we have a question where we need to isolate a variable $u$ in terms of another variable $q$. Currently, I am stuck with a rather difficult expression to simplify or at least I can't see what the trick is (if there is one). Here is the said expression:



$$log_2big(q^q(1-q)^{1-q}big) = log_2bigg((frac{qu}{1-qu})^qbigg)$$



Being perfectly honest, I know that the $log_2$ simplifies on each side but having $u$ in both the numerator and denominator of the fraction which has an overall exponent, I'm quite out of my comfort zone mathematically.



Thanks in advance!






calculus exponentiation coding-theory







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  • $begingroup$
    What is $u$? It doesn't appear in the left hand side
    $endgroup$
    – pwerth
    Jan 24 at 19:47










  • $begingroup$
    That was one of my problems ^^
    $endgroup$
    – Max Michel
    Jan 24 at 19:58
















0












$begingroup$


I'm currently revising an exam about channel coding in telecommunications and we have a question where we need to isolate a variable $u$ in terms of another variable $q$. Currently, I am stuck with a rather difficult expression to simplify or at least I can't see what the trick is (if there is one). Here is the said expression:



$$log_2big(q^q(1-q)^{1-q}big) = log_2bigg((frac{qu}{1-qu})^qbigg)$$



Being perfectly honest, I know that the $log_2$ simplifies on each side but having $u$ in both the numerator and denominator of the fraction which has an overall exponent, I'm quite out of my comfort zone mathematically.



Thanks in advance!






calculus exponentiation coding-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is $u$? It doesn't appear in the left hand side
    $endgroup$
    – pwerth
    Jan 24 at 19:47










  • $begingroup$
    That was one of my problems ^^
    $endgroup$
    – Max Michel
    Jan 24 at 19:58














0












0








0





$begingroup$


I'm currently revising an exam about channel coding in telecommunications and we have a question where we need to isolate a variable $u$ in terms of another variable $q$. Currently, I am stuck with a rather difficult expression to simplify or at least I can't see what the trick is (if there is one). Here is the said expression:



$$log_2big(q^q(1-q)^{1-q}big) = log_2bigg((frac{qu}{1-qu})^qbigg)$$



Being perfectly honest, I know that the $log_2$ simplifies on each side but having $u$ in both the numerator and denominator of the fraction which has an overall exponent, I'm quite out of my comfort zone mathematically.



Thanks in advance!






calculus exponentiation coding-theory







share|cite|improve this question









$endgroup$




I'm currently revising an exam about channel coding in telecommunications and we have a question where we need to isolate a variable $u$ in terms of another variable $q$. Currently, I am stuck with a rather difficult expression to simplify or at least I can't see what the trick is (if there is one). Here is the said expression:



$$log_2big(q^q(1-q)^{1-q}big) = log_2bigg((frac{qu}{1-qu})^qbigg)$$



Being perfectly honest, I know that the $log_2$ simplifies on each side but having $u$ in both the numerator and denominator of the fraction which has an overall exponent, I'm quite out of my comfort zone mathematically.



Thanks in advance!






calculus exponentiation coding-theory




calculus exponentiation coding-theory






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 19:42









Max MichelMax Michel

32




32












  • $begingroup$
    What is $u$? It doesn't appear in the left hand side
    $endgroup$
    – pwerth
    Jan 24 at 19:47










  • $begingroup$
    That was one of my problems ^^
    $endgroup$
    – Max Michel
    Jan 24 at 19:58


















  • $begingroup$
    What is $u$? It doesn't appear in the left hand side
    $endgroup$
    – pwerth
    Jan 24 at 19:47










  • $begingroup$
    That was one of my problems ^^
    $endgroup$
    – Max Michel
    Jan 24 at 19:58
















$begingroup$
What is $u$? It doesn't appear in the left hand side
$endgroup$
– pwerth
Jan 24 at 19:47




$begingroup$
What is $u$? It doesn't appear in the left hand side
$endgroup$
– pwerth
Jan 24 at 19:47












$begingroup$
That was one of my problems ^^
$endgroup$
– Max Michel
Jan 24 at 19:58




$begingroup$
That was one of my problems ^^
$endgroup$
– Max Michel
Jan 24 at 19:58










1 Answer
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$begingroup$

Since we have on both sides the base $2$ we get
$$q^q(1-q)^{1-q}=left(frac {qu}{1-qu}right)^q$$ Let's say $$A=q^q(1-q)^{1-q}$$ then $$a^{1/q}=frac{qu}{1-qu}$$ so $$A^{1/q}-qA^{1/q}=qu$$ and $$A^{1/q}-quA^{1/q}=qu$$ so $$A^{1/q}=u(q+qA^{1/q})$$ and we obtain
$$u=frac{A^{1/q}}{q+qA^{1/q}}$$






share|cite|improve this answer









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    $begingroup$

    Since we have on both sides the base $2$ we get
    $$q^q(1-q)^{1-q}=left(frac {qu}{1-qu}right)^q$$ Let's say $$A=q^q(1-q)^{1-q}$$ then $$a^{1/q}=frac{qu}{1-qu}$$ so $$A^{1/q}-qA^{1/q}=qu$$ and $$A^{1/q}-quA^{1/q}=qu$$ so $$A^{1/q}=u(q+qA^{1/q})$$ and we obtain
    $$u=frac{A^{1/q}}{q+qA^{1/q}}$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since we have on both sides the base $2$ we get
      $$q^q(1-q)^{1-q}=left(frac {qu}{1-qu}right)^q$$ Let's say $$A=q^q(1-q)^{1-q}$$ then $$a^{1/q}=frac{qu}{1-qu}$$ so $$A^{1/q}-qA^{1/q}=qu$$ and $$A^{1/q}-quA^{1/q}=qu$$ so $$A^{1/q}=u(q+qA^{1/q})$$ and we obtain
      $$u=frac{A^{1/q}}{q+qA^{1/q}}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since we have on both sides the base $2$ we get
        $$q^q(1-q)^{1-q}=left(frac {qu}{1-qu}right)^q$$ Let's say $$A=q^q(1-q)^{1-q}$$ then $$a^{1/q}=frac{qu}{1-qu}$$ so $$A^{1/q}-qA^{1/q}=qu$$ and $$A^{1/q}-quA^{1/q}=qu$$ so $$A^{1/q}=u(q+qA^{1/q})$$ and we obtain
        $$u=frac{A^{1/q}}{q+qA^{1/q}}$$






        share|cite|improve this answer









        $endgroup$



        Since we have on both sides the base $2$ we get
        $$q^q(1-q)^{1-q}=left(frac {qu}{1-qu}right)^q$$ Let's say $$A=q^q(1-q)^{1-q}$$ then $$a^{1/q}=frac{qu}{1-qu}$$ so $$A^{1/q}-qA^{1/q}=qu$$ and $$A^{1/q}-quA^{1/q}=qu$$ so $$A^{1/q}=u(q+qA^{1/q})$$ and we obtain
        $$u=frac{A^{1/q}}{q+qA^{1/q}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 24 at 19:52









        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

        77.9k42866




        77.9k42866






























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