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Question Number 199389 by depressiveshrek last updated on 02/Nov/23

$${\log }_{12}60=?$$$${\log }_{6}30=a$$$${\log }_{15}24=b$$

Answered by cortano12 last updated on 03/Nov/23

$$\log {}_{12}\left(60\right)=\frac{\log {}_6(6.5.2)}{\log {}_6(6.2)}$$$$=\frac{2\mathrm{ab}+2\mathrm{a}-1}{\mathrm{ab}+\mathrm{b}+1}$$

Commented by depressiveshrek last updated on 03/Nov/23

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{how}\mathrm{you}\mathrm{got}\mathrm{there}...$$

Answered by mr W last updated on 04/Nov/23

$${\log }_{15}24=\frac{{\log }_{6}24}{{\log }_{6}15}=\frac{{\log }_{6}6+{2\log }_{6}2}{{\log }_{6}30-{\log }_{6}2}$$$$=\frac{1+{2\log }_{6}2}{{\log }_{6}30-{\log }_{6}2}$$$$=\frac{1+{2\log }_{6}2}{a-{\log }_{6}2}=b$$$$\Rightarrow {\log }_{6}2=\frac{ab-1}{2+b}$$$$$$$${\log }_{12}60$$$$=\frac{{\log }_{6}30+{\log }_{6}2}{{\log }_{6}6+{\log }_{6}2}$$$$=\frac{{\log }_{6}30+{\log }_{6}2}{1+{\log }_{6}2}$$$$=\frac{a+\frac{ab-1}{2+b}}{1+\frac{ab-1}{2+b}}$$$$=\frac{2a(b+1)-1}{(a+1)b+1}✓$$

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