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One student sent me this question:

If $\log_35=a$ and $\log_54=b$, what is $\log_{60}70$?

Question asks the value of $\log_{60}70$ in terms of $a$ and $b$. Equations for $a$ and $b$ involved $2$, $3$ and $5$. So I am not sure how to deal with $7$ in $\log_{60}70$ by using logarithm identities. I may be missing something very obvious.

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  • $\begingroup$ Maybe the change of base formula for logarithms will be useful! $\endgroup$
    – ET-phone-homology
    Commented May 1, 2016 at 7:24
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    $\begingroup$ @Smath but that is exactly what the question is about! $\endgroup$
    – N.S.JOHN
    Commented May 1, 2016 at 7:29
  • $\begingroup$ @Smath, as i stated in the question, i cannot get rid of $7$ in that way. $\endgroup$
    – Alistair
    Commented May 1, 2016 at 7:33
  • $\begingroup$ I think the answer expected will have log(7) as a constant term in it. $\endgroup$
    – user1952500
    Commented May 1, 2016 at 8:19

1 Answer 1

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If there is an algebraic function $f$ with $$ \log_{60}70 =f(a,b)= f\left(\frac{\log 5}{\log 3}, \frac{\log 4}{\log 5}\right)$$ then since $$ \log_{60}70 = \frac{\log 70}{\log 60} = \frac{\log 7+\log 2+\log 5}{2\log 2+\log 3+\log 5},$$ we would have $$ \log 7 = f\left(\frac{\log 5}{\log 3}, \frac{\log 4}{\log 5}\right)(2\log 2+\log 3+\log 5) - \log 2 - \log 5 $$ so $\log 7$ would be algebraic in $\log 2$, $\log 3$, and $\log 5$. But according to Schanuel's conjecture, the logarithms of the primes are algebraically independent.

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  • $\begingroup$ This solves it. $\endgroup$
    – N.S.JOHN
    Commented May 1, 2016 at 8:39

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