Can someone help me to find $\log_{12}{60}$ if there is given $\log_{6}30$=a and $\log_{15}24$=b?
Find $\log_{12}{60}$ if there is given $\log_{6}30$=a and $\log_{15}24$=b
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$\begingroup$
logarithms
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1Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – 2017-01-22
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1The notation log(a,b) is unclear to me. – 2017-01-22
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0Did you mean$$\log_{12}(60)$$?$\quad$ – 2017-01-22
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0how is this to be read $$\log_{12} 60$$? – 2017-01-22
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0Yeah I didn't know how to write that – 2017-01-22
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0Please see here: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-01-22
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0$\log_{12}(60)=\log_{12}(2^{a-1})-a$ – 2017-01-22
1 Answers
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Hints: Fill in details
$$\begin{align*} &\log_{12}60=\frac{\log_660}{\log_612}=\frac{\log_65}{1+\log_62}+1\\{}\\ &\log_{15}24=\frac{\log_624}{\log_615}=\frac{1+2\log_62}{1+\log_65-\log_62}\\{}\\ &\log_630=1+\log_65\end{align*}$$
And now just substitute (why and how?):
$$\log_65=a-1\;,\;\;b(a-\log_62)=1+2\log_62\implies (b+2)\log_62=ab-1\;\ldots$$