Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

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

If    log_6 30   =   a,         log_(15) 24   =   b,           evaluate:   log_(12) 60

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

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

nice!

$$\mathrm{nice}! \\ $$

Answered by bramlexs22 last updated on 20/Jul/21

  If  { ((log _6 (30)=a)),((log _(15) (24)=b)) :} Evaluate log _(12) (60).  ≡ solution ≡  (•) a=log _6 (6×5)=1+log _6 (5)   ⇒ log _6 (5)=a−1  (•)b= log_(15) (24)=((log _6 (24))/(log _6 (15)))=((1+2log_6 (2) )/(a−1+log _6 (3)))   ⇒ab−b+b(1−log _6 (2))=1+2log _6 (2)  ⇒ab−1=(2+b)log _6 (2)  ⇒log _6 (2)=((ab−1)/(b+2))  (•)log _(12) (60)=((1+log _6 (5)+log _6 (2))/(1+log _6 (2)))  = ((1+a−1+log _6 (2))/(1+log _6 (2)))=((a+log _6 (2))/(1+log _6 (2)))  =((a+((ab−1)/(b+2)))/(1+((ab−1)/(b+2)))) = ((2ab+2a−1)/(ab+b+1)) .

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

Commented by Tawa11 last updated on 20/Jul/21

Thanks sir. God bless you.

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com