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Question Number 126430 by TITA last updated on 20/Dec/20

$$\log {}_6^{15}=\mathrm{a}\log {}_{12}^{18}=\mathrm{b}\mathrm{find}\log {}_{25}^{24}$$$$\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}$$

Commented by TITA last updated on 20/Dec/20

$$\mathrm{please}\mathrm{help}$$

Answered by mr W last updated on 20/Dec/20

$$let\log 2=x,\log 3=y$$$${\log }_{6}15=\frac{\log 15}{\log 6}=\frac{\log 3+\log 10-\log 2}{\log 3+\log 2}$$$$=\frac{\log 3+1-\log 2}{\log 3+\log 2}=\frac{y+1-x}{y+x}=a$$$$\Rightarrow (1+a)x-(1-a)y=1...\left(i\right)$$$${\log }_{12}18=\frac{\log 18}{\log 12}=\frac{\log 2+2\log 3}{\log 3+2\log 2}$$$$=\frac{x+2y}{y+2x}=b$$$$\Rightarrow (1-2b)x+(2-b)y=0...\left(ii\right)$$$$from\left(i\right)and\left(ii\right)weget$$$$x=\frac{b-2}{(1+a)(b-2)-(1-a)(1-2b)}$$$$y=\frac{1-2b}{(1+a)(b-2)-(1-a)(1-2b)}$$$$$$$${\log }_{25}24=\frac{\log 24}{\log 25}=\frac{\log 3+3\log 2}{2-2\log 2}=\frac{3x+y}{2(1-x)}$$$$=\frac{3(b-2)+(1-2b)}{2[(1+a)(b-2)-(1-a)(1-2b)-b+2]}$$$$\Rightarrow \frac{b-5}{2(2b-a-ab-1)}$$

Commented by Dwaipayan Shikari last updated on 20/Dec/20

$$Wehave9°CinKolkata$$

Commented by MJS_new last updated on 20/Dec/20

$$\mathrm{lol}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{baking}\mathrm{cookies}\mathrm{so}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{as}\mathrm{fast}\mathrm{as}\mathrm{you}...$$

Commented by mr W last updated on 20/Dec/20

$$i^{\prime }mjusteatingcookieswithGl\ddot{uhwein},$$$$soimustbefaster...$$

Commented by MJS_new last updated on 20/Dec/20

$$\mathrm{I}\mathrm{made}\mathrm{Elisenlebkuchen},\mathrm{Vanillekipferl}\mathrm{\&}$$$$\mathrm{Zimtsterne}...\mathrm{season}\mathrm{greetings}\mathrm{from}\mathrm{Vienna}!$$

Commented by Dwaipayan Shikari last updated on 20/Dec/20

$$WhatistheweatherinViennasir?Maybecold$$

Commented by Ar Brandon last updated on 20/Dec/20

����Mr W and Mr MJS

Commented by MJS_new last updated on 20/Dec/20

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{now}5:40\mathrm{p}.\mathrm{m}.,\mathrm{sunset}\mathrm{was}4:02\mathrm{p}.\mathrm{m}.$$$$\mathrm{we}\mathrm{have}+3°\mathrm{C}\mathrm{and}\mathrm{mist}/\mathrm{light}\mathrm{rain}$$

Commented by talminator2856791 last updated on 22/Dec/20

$$\mathrm{in}\mathrm{south}\mathrm{africa}\mathrm{it}\mathrm{is}\mathrm{around}20°\mathrm{in}\mathrm{morning}\mathrm{now}$$

Answered by MJS_new last updated on 20/Dec/20

$$a=\frac{\ln 15}{\ln 6}=\frac{\ln 3+\ln 5}{\ln 2+\ln 3}$$$$b=\frac{\ln 18}{\ln 12}=\frac{\ln 2+2\ln 3}{2\ln 2+\ln 3}$$$$\mathrm{let}\ln 2=p;\ln 3=q;\ln 5=r$$$$a=\frac{q+r}{p+q}\Rightarrow p=\frac{-qa+q+r}{a}$$$$b=\frac{p+2q}{2p+q}=\frac{-qa-q-r}{qa-2q-2r}\Rightarrow q=\frac{(2b-1)r}{ab+a-2b+1}$$$$\Rightarrow p=\frac{(2-b)r}{ab+a-2b+1}$$$$\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{know}\frac{\ln 24}{\ln 25}=\frac{3\ln 2+\ln 2}{2\ln 5}=\frac{3p+q}{2r}$$$$\mathrm{and}\mathrm{inserting}\mathrm{we}\mathrm{get}$$$$\frac{5-b}{2(ab+a-2b+1)}$$

Answered by Dwaipayan Shikari last updated on 20/Dec/20

$$\frac{log15}{log6}=a\Rightarrow log\left(5\right)+log\left(3\right)=alog\left(3\right)+alog\left(2\right)\to \left(a\right)$$$$\frac{log\left(18\right)}{log\left(12\right)}=b\Rightarrow 2log\left(3\right)+log\left(2\right)=blog\left(3\right)+2blog\left(2\right)\Rightarrow \frac{log3}{log2}=\frac{2b-1}{2-b}$$$$From\left(a\right)log\left(5\right)=(a-1)log3+alog\left(2\right)=log\left(2\right)(\frac{(a-1)(2b-1)}{2-b}+a)$$$$\frac{log\left(8\right)+log\left(3\right)}{2log\left(5\right)}=\frac{3log\left(2\right)+log\left(2\right)\left(\frac{2b-1}{2-b}\right)}{2log\left(2\right)\left(\frac{2ab-2b-a+1+2a-ab}{2-b}\right)}=\frac{5-b}{2(ab+a-2b+1)}$$

Commented by Dwaipayan Shikari last updated on 20/Dec/20

$$Slowestpossibleanswer$$😁

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