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Question Number 92010 by Rio Michael last updated on 04/May/20

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

Answered by jagoll last updated on 04/May/20

$${\log }_{12}\left(60\right)=\frac{{\log }_6\left(30\right)+{\log }_6\left(2\right)}{1+{\log }_6\left(2\right)}$$$$=\frac{\mathrm{a}+{\log }_6\left(2\right)}{1+{\log }_6\left(2\right)}=\frac{\mathrm{a}+\left(\frac{\mathrm{a}-\mathrm{b}}{2\mathrm{b}+1}\right)}{1+\left(\frac{\mathrm{a}-\mathrm{b}}{2\mathrm{b}+1}\right)}$$$$=\frac{\mathrm{a}(2\mathrm{b}+1)+\mathrm{a}-\mathrm{b}}{2\mathrm{b}+1+\mathrm{a}-\mathrm{b}}=\frac{2\mathrm{a}+2\mathrm{ab}-\mathrm{b}}{\mathrm{a}+\mathrm{b}+1}$$$$\left(1\right)\log {}_{24}\left(15\right)=\frac{\log {}_6\left(30\right)-\log {}_6\left(2\right)}{1+2\log {}_6\left(2\right)}=\mathrm{b}$$$$\mathrm{a}-\log {}_6\left(2\right)=\mathrm{b}+2\mathrm{b}\log {}_6\left(2\right)$$$$(2\mathrm{b}+1)\log {}_6\left(2\right)=\mathrm{a}-\mathrm{b}$$$$\log {}_6\left(2\right)=\frac{\mathrm{a}-\mathrm{b}}{2\mathrm{b}+1}$$

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