Denesting Logarithmic expressions \(\displaystyle{{\log}_{{7}}{\left({{\log}_{{2}}{\left({3}\right)}}\right)}}+{{\log}_{{7}}{\left({{\log}_{{5}}{\left({6}\right)}}\right)}}+{{\log}_{{7}}{\left({{\log}_{{{11}}}{\left(\frac{{1}}{{2}}\right)}}\right)}}={{\log}_{{7}}{\left(-{1}\right)}}+{{\log}_{{7}}{\left({{\log}_{{5}}{\left({3}\right)}}\right)}}+{{\log}_{{7}}{\left({{\log}_{{{11}}}{\left({6}\right)}}\right)}}\) This can only be simplified

dotzis16xd

dotzis16xd

Answered question

2022-03-25

Denesting Logarithmic expressions
log7(log2(3))+log7(log5(6))+log7(log11(12))=log7(1)+log7(log5(3))+log7(log11(6))
This can only be simplified by using the sum to product rule and noticing that of the n! possible permutations that 12 is a power of 2. Because N is small and the relationship between 2 to 12 is simple this can be solved trivially. How does one go about simplifying much more complex logarithmic identities in polynomial time?

Answer & Explanation

Mason Knight

Mason Knight

Beginner2022-03-26Added 11 answers

If I'm understanding what you're looking for (and that -1 in parentheses on the right-hand side is illegal, as we'll come to below), your expression works because of the "Change-of-Base" Formula, logb(x)=loga(x)loga(b)
So, as you say, because the sum of logarithms gives the logarithm of a product, you can carry on permutations of the sort you have here for any number of terms:
loga[logb1(x1)]+loga[logb2(x2)]++loga[logbn(xn)]
=loga[logb1(x1)logb2(x2)logbn(xn)]
=loga[ln(x1)ln(b1)ln(x2)ln(b2)ln(xn)ln(bn)]=loga[ln(x1)ln(x2)ln(xn)ln(b1)ln(b2)ln(bn)].
You can see from this that you are then free to permute the factors in the product in either the numerator or the denominator to create any combination of numbers xi and logarithmic bases bi you wish.
OK, having said that, I see what is going on in your equation. There is a domain issue here in order to apply this result. The nested value loga(logb(x)) must be defined for every term in the sum. Thus, you cannot have any terms where any of the xi are less than or equal to 1; otherwise, you will have a term with logbi(x)0 for which you will not be able to evaluate loga(logbi(x))

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