Given-log-2-10-a-a-1-log-3-5-1-b-find-the-value-of-1-log-12-15-

Question Number 120380 by bramlexs22 last updated on 31/Oct/20

$${Given}\:\begin{cases}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{10}\right)=\frac{{a}}{{a}−\mathrm{1}}}\\{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)=\frac{\mathrm{1}}{{b}}}\end{cases} \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{1}+\mathrm{log}\:_{\mathrm{12}} \left(\mathrm{15}\right)\:? \\ $$

Answered by FelipeLz last updated on 31/Oct/20

log_2 10 = 1+((ln 5)/(ln 2)) = (a/(a−1)) ⇒ ln 2 = (a−1)ln 5 log_3 5 = ((ln 5)/(ln 3)) = (1/b) ⇒ ln 3 = bln 5 1+log_(12) 15 = 1+((ln 3 + ln 5)/(ln 3 + 2ln 2)) = 1+((bln 5 + ln 5)/(bln 5 + 2(a−1)ln 5)) = 1+((b+1)/(b+2(a−1))) = ((2(a+b)−1)/(2(a−1)+b))

$$\mathrm{log}_{\mathrm{2}} \:\mathrm{10}\:=\:\mathrm{1}+\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{ln}\:\mathrm{2}}\:=\:\frac{{a}}{{a}−\mathrm{1}}\:\Rightarrow\:\mathrm{ln}\:\mathrm{2}\:=\:\left({a}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{5} \\ $$$$\mathrm{log}_{\mathrm{3}} \:\mathrm{5}\:=\:\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{ln}\:\mathrm{3}}\:=\:\frac{\mathrm{1}}{{b}}\:\Rightarrow\:\mathrm{ln}\:\mathrm{3}\:=\:{b}\mathrm{ln}\:\mathrm{5} \\ $$$$\mathrm{1}+\mathrm{log}_{\mathrm{12}} \:\mathrm{15}\:=\:\mathrm{1}+\frac{\mathrm{ln}\:\mathrm{3}\:+\:\mathrm{ln}\:\mathrm{5}}{\mathrm{ln}\:\mathrm{3}\:+\:\mathrm{2ln}\:\mathrm{2}}\:=\:\mathrm{1}+\frac{{b}\mathrm{ln}\:\mathrm{5}\:+\:\mathrm{ln}\:\mathrm{5}}{{b}\mathrm{ln}\:\mathrm{5}\:+\:\mathrm{2}\left({a}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{5}}\:=\:\mathrm{1}+\frac{{b}+\mathrm{1}}{{b}+\mathrm{2}\left({a}−\mathrm{1}\right)}\:=\:\frac{\mathrm{2}\left({a}+{b}\right)−\mathrm{1}}{\mathrm{2}\left({a}−\mathrm{1}\right)+{b}} \\ $$

Answered by john santu last updated on 31/Oct/20

{ ((log _2 (5)+1 = (a/(a−1)) ⇒log _2 (5)=((a−a+1)/(a−1)) = (1/(a−1)))),((log _3 (5)=(1/b))) :} ⇒1+((log _5 (3)+1)/(log _5 (3)+2log _5 (2))) = 1+((b+1)/(b+2a−2)) ⇒ ((2a+2b−1)/(2a+b−2))

$$\begin{cases}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{5}\right)+\mathrm{1}\:=\:\frac{{a}}{{a}−\mathrm{1}}\:\Rightarrow\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{5}\right)=\frac{{a}−{a}+\mathrm{1}}{{a}−\mathrm{1}}\:=\:\frac{\mathrm{1}}{{a}−\mathrm{1}}}\\{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}\right)=\frac{\mathrm{1}}{{b}}}\end{cases} \\ $$$$\Rightarrow\mathrm{1}+\frac{\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right)+\mathrm{1}}{\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right)+\mathrm{2log}\:_{\mathrm{5}} \left(\mathrm{2}\right)}\:=\:\mathrm{1}+\frac{{b}+\mathrm{1}}{{b}+\mathrm{2}{a}−\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}{a}+\mathrm{2}{b}−\mathrm{1}}{\mathrm{2}{a}+{b}−\mathrm{2}} \\ $$

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