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Assuming $a=\log 2$ and $b=\log 3$ (log is the base 10 logarithm). I have to find $\log_5 288$. How can I do this?

I've tried transforming $\log2$ to $\frac{\log_5 2}{\log_5 10}$ and same for $b$. Then it's $$\frac{5\log_5 2+2\log_5 3}{\log_5 10}=\frac{\log_5 288}{\log_5 10}=5\log2+2\log5=5a+3b$$ Is that correct?

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3 Answers 3

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HINT:

Using $$\log_ab=\frac{\log_cb}{\log_ca},$$

$$\log_5{288}=\log_5(2^53^2)=5\log_52+2\log_53=\frac{5\log_{10}2+2\log_{10}3}{\log_{10}5}$$

Now $\displaystyle 1=\log_{10}{10}=\log_{10}2+\log_{10}5\iff \log_{10}5=1-\log_{10}2=\cdots$

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$$\log 2=\frac{\log_5 2}{\log_5 10}=\frac{\log_5 2}{\log_5 2+\log_5 5}=\frac{\log_5 2}{\log_5 2+1}\qquad \log 3=\frac{\log_5 3}{\log_5 10}=\frac{\log_5 3}{\log_5 2+1} $$ Hence $$\log_5 2=\frac{\log 2}{1-\log2}=\frac{a}{1-a}\qquad \log_5 3= \log 3 (\log_5 2+1)=b\left(\frac{a}{1-a}+1\right)=\frac{b}{1-a}\ .$$ Hence $$\log_5 288=\log_5 2^5\cdot3^2=5\log_5 2+2\log_5 3=5\frac{a}{1-a}+2\frac{b}{1-a}=\frac{5a+2b}{1-a}\ .$$

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\begin{align} \log_5 288 & = \frac{\log_{10} 288}{\log_{10} 5} = \frac{\log_{10}(2\cdot2\cdot2\cdot2\cdot2\cdot3\cdot3)}{\log_{10} (10/2)} \\[15pt] & = \frac{\log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}3 + \log_{10}3 + }{\log_{10} 10 - \log_{10} 2} \\[15pt] & = \frac{\log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}3 + \log_{10}3 + }{1 - \log_{10} 2} \end{align}

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