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it is my exercise to compute the numerical value of $x$ and $y$, where $$x:=\log_{20}(100)+\log_{100}(20),\\ y:=\log_{\frac{1}{2}}(70)+\log_{\frac{1}{10}}(200)+\log_{\frac{3}{2}}\frac{1}{100}.$$

In order to do so, I tried to simplify the expressions in various ways (by using the usual rules of logarithm), but I can not do it in a satisfactory way, such that the values are obvious. Is there any trick you can tell me which might work?

Let my show you, for example, what I tried to find $x$ explicitly. First I observed that the terms $\log_{20}(100)$ and $\log_{100}(20)$ do not seem to be computable individually. As I said before, I tried then to manipulate it in various ways, but I always get stuck, e.g.:

$$x= \log_{20}(100) + \frac{1}{\log_{20}(100)}=?\\ x= 2 \log_{20}(5)+ 2\log_{20}(2)+ 2\log_{100}(2)+ \log_{100}(5) = ?$$ etc.

I would appreciate your help very much!

Best

Oliver Watt
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  • For the 1st eqn, note that $(20)^a = 100 \implies 100^{(1/a)} = 20.$ – user2661923 Jan 18 '21 at 09:30
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    @user2661923 How will you find out the value of $\alpha$? –  Jan 18 '21 at 09:31
  • @Forester Since $\log_{10}~2$ is transcendental, I would regard it as reasonable to set $r = \log_{10}~2$ and then express the computation in terms of $r$. – user2661923 Jan 18 '21 at 09:32
  • Re my first comment, $a$ is clearly a transcendental number, so the only real issue is whether there is any dramatic simplification inherent in $\left(a + \frac{1}{a}\right).$ – user2661923 Jan 18 '21 at 09:36
  • Are you supposed to compute the numerical value accurate to a fixed set of decimal places (e.g. accurate to 3 decimal places)? My calculator gives $\log_{10}~20 \approx 1.30103 \implies \log_{100}~20 \approx \frac{1.30103}{2}.$ – user2661923 Jan 18 '21 at 09:48
  • you can express $x,y$ in terms of $\log _z2, \log_z 3, \log_z 5, \log_z 7$, where $z$ any arbitrary positive number. – Z Ahmed Jan 18 '21 at 09:49
  • @ZAhmed True, and normally I would advise setting $z = 10,$ and go from there. However, consider: "it is my exercise to compute the numerical value of $x$ and $y$...". What exactly is the problem composer's intent behind compute the numerical value...? – user2661923 Jan 18 '21 at 09:52
  • May be you may have use $\log_{10} 2=0.3010, \log_{10} 3=0.4771,...$ etc. I don't see any thing else here. – Z Ahmed Jan 18 '21 at 09:54

1 Answers1

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There is a formula to change the base of logarithms $$\log_a b=\frac{\log_c b}{\log_c a}$$ for instance $$\log_{20} 100+\log_{100} 20=\frac{\log_{100} 100}{\log_{100} 20}+\log_{100} 20=\frac{1}{\log_{100} 20}+\log_{100} 20$$ Note that $$\log_{100}x=\frac{\log_{10}x}{\log_{10}100}=\frac{1}{2}\log_{10}x$$

Raffaele
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