Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

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

Given  { ((log _2 (10)=(a/(a−1)))),((log _3 (5)=(1/b))) :}  find the value of 1+log _(12) (15) ?

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

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))

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com