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I am very interested in mathematics, however, finding nowhere near wanted information in school sometimes I go and learn something by myself. Just like this time. I decided to learn more about logarithms as they always picked my interest, even thought it's a year or two more advanced than I should be. So, I found an article at Brilliant.org and decided to learn everything they have to offer (I love the difficulty of problems on that side). But I stumbled across one huge problem: after reading about basic properties of logarithms and looking through all of the examples I was left with no practice! I know that I could find a few logarithm based problems on the website, however, before I dwell into advanced problems, I want to be able to solve average ones. I am looking for the kind of exercises that the website showed as examples ( Simplify $${2\log_4{\sqrt{5}}+\frac{1}{2}\log_2{625}-\log_2{\frac{1}{5}}}$$ and this kind of examples that test pretty much every aspect understanding the very basics). Thanks!

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$2\log_4(\sqrt{5})=\log_4(5)=\frac{\log_2(5)}{\log_2(4)}=\log_2\sqrt{5}$ (change of base rule).
Now $\frac{1}{2}\log_2(625)=\log_2(\sqrt{625})=\log_2(25)$ now $\log a-\log b=\log\frac{a}{b},\, \log a+\log b=\log(ab)$. So now $\log_2(\sqrt{5})+\log_2(25)+\log_2\left(\frac{1}{2}\right)$ simplifies to $\log_2\left(\frac{\sqrt{5}\cdot 25}{1/5}\right)=\log_2(125\sqrt{5})$.

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  • $\begingroup$ The question had $2log_4(\sqrt{5})$ before edit. $\endgroup$ Dec 12, 2015 at 14:21
  • $\begingroup$ @AmeetSharma It has already been rolled back. @ Archis Welankar If you think the OP has made a mistake of this kind then it is probably wise to ask for clarification first. $\endgroup$
    – gebruiker
    Dec 12, 2015 at 14:25
  • $\begingroup$ Thank you for your work, however, I am searching for more exercises of such, sorry that I didn't state so clearly on the post (and as Ameet stated, it's $log_4$). $\endgroup$
    – Zyberg
    Dec 12, 2015 at 14:26
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Sorry for that : I only use natural logarithms. So, from the definition $$\log_a(b)=\frac{\log (b)}{\log (a)}$$ we have $$t_1=\log_4(\sqrt{5})=\frac{\log (5)}{2 \log (4)}=\frac{\log (5)}{4 \log (2)}$$ $$t_2=\log_2(625)=\frac{\log (625)}{\log (2)}$$ $$t_3=\log_2(\frac{1}{5})=-\frac{\log (5)}{\log (2)}$$ $$2t_1+\frac 12 t_2-t_3=\frac{7 \log (5)}{2\log (2)}=\frac{ \log (5^{7/2})}{\log (2)}=\frac{ \log (125 \sqrt5)}{\log (2)}=\log_2(125 \sqrt5)$$ This is a (at least my) general approach : in the specific case of the post, things are simpler because only $\log_4(.)$ and $\log_2(.)$ are used and these are very cloçsely related to each other.

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