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I don't know how to compare logarithmic expressions.

Which is greater?

$ \log_23 \ or \log_\frac{1}{2}5 \ $

I can, of course compare using calculator but I need the steps to compare using pen and paper.

I need a way that works on any question. so as to compare this kind of expressions e.g.

$ \log_711 \ and \log_85 \ $

kingW3
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Fghj
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  • Here on MSE we prefer if you show your own work before just asking for the solution to an exercise. In particular, you could provide more context e.g. where you encountered this problem. Also, have you considered a calculator ? – rah4927 Jul 02 '17 at 09:54
  • hiNT:$$\log_\frac{1}{2}5=\log_{2^{-1}}5=\frac{1}{-1}\log_25$$ – Khosrotash Jul 02 '17 at 09:55
  • Also please don't change the question, if you have a new question ask it. – kingW3 Jul 02 '17 at 10:30

2 Answers2

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Although, we can compare it using the property of logarithms, that allows us to split the base and the domain, but I would prefer to post a logical solution.

Any positive power raised to $\frac 12$ (or any other positive number less than $1$) will result in a smaller no. than itself. Therefore, if you want a number greater than $\frac 12$, you'll have to raise negative power.

Also, If we want a greater number than $2$, we'll have to raise it to a positive power for sure.

Hence, $$\log_2 3 > \log_{\frac 12}5$$

Jaideep Khare
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  • See, will this work for the updated question? – Fghj Jul 02 '17 at 10:09
  • @RaviPrakash For numbers greater than 1 : If the base is smaller than domain -- result will be greater than 1, and if the base is greater than domain, than the result will be less than 1. – Jaideep Khare Jul 02 '17 at 10:40
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Write $x = \log_{2}3$ and $y = \log_{\frac{1}{2}}5$. So that $2^{x} = 3$ and $2^{-y} = 5$. We get $2^{x - y} = 15$. So that $x - y = \log_{2}15 > 0$, which shows that $x > y$.

Daniel Akech Thiong
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