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Question Number 147334 by Tawa11 last updated on 19/Jul/21

$$\mathrm{If}{\log }_{6}30=\mathrm{a},{\log }_{15}24=\mathrm{b},\mathrm{evaluate}:{\log }_{12}60$$

Commented by otchereabdullai@gmail.com last updated on 20/Jul/21

$$\mathrm{nice}!$$

Answered by bramlexs22 last updated on 20/Jul/21

$$\mathrm{If}\{\begin{array}{l}\log {}_6\left(30\right)=\mathrm{a}\\ \log {}_{15}\left(24\right)=\mathrm{b}\end{array}\mathrm{Evaluate}\log {}_{12}\left(60\right).$$$$\equiv \mathrm{solution}\equiv$$$$(\bullet )\mathrm{a}=\log {}_6\left(6\times 5\right)=1+\log {}_6\left(5\right)$$$$\Rightarrow \log {}_6\left(5\right)=\mathrm{a}-1$$$$(\bullet )\mathrm{b}={\log }_{15}\left(24\right)=\frac{\log {}_6\left(24\right)}{\log {}_6\left(15\right)}=\frac{1+{2\log }_6\left(2\right)}{\mathrm{a}-1+\log {}_6\left(3\right)}$$$$\Rightarrow \mathrm{ab}-\mathrm{b}+\mathrm{b}(1-\log {}_6\left(2\right))=1+2\log {}_6\left(2\right)$$$$\Rightarrow \mathrm{ab}-1=(2+\mathrm{b})\log {}_6\left(2\right)$$$$\Rightarrow \log {}_6\left(2\right)=\frac{\mathrm{ab}-1}{\mathrm{b}+2}$$$$(\bullet )\log {}_{12}\left(60\right)=\frac{1+\log {}_6\left(5\right)+\log {}_6\left(2\right)}{1+\log {}_6\left(2\right)}$$$$=\frac{1+\mathrm{a}-1+\log {}_6\left(2\right)}{1+\log {}_6\left(2\right)}=\frac{\mathrm{a}+\log {}_6\left(2\right)}{1+\log {}_6\left(2\right)}$$$$=\frac{\mathrm{a}+\frac{\mathrm{ab}-1}{\mathrm{b}+2}}{1+\frac{\mathrm{ab}-1}{\mathrm{b}+2}}=\frac{2\mathrm{ab}+2\mathrm{a}-1}{\mathrm{ab}+\mathrm{b}+1}.$$

Commented by Tawa11 last updated on 20/Jul/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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