Menu Close

log-6-15-a-log-12-18-b-find-log-25-24-in-terms-of-a-and-b-

Tinku Tara 0 Comments
Pin




Question Number 126430 by TITA last updated on 20/Dec/20
log _6^(15) =a log _(12)^(18) =b find log _(25)^(24) in terms of a and b
$$\mathrm{log}\:_{\mathrm{6}} ^{\mathrm{15}} =\mathrm{a}\:\:\:\mathrm{log}\:_{\mathrm{12}} ^{\mathrm{18}} =\mathrm{b}\:\:\mathrm{find}\:\mathrm{log}\:_{\mathrm{25}} ^{\mathrm{24}} \\ $$$$\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b} \\ $$
Commented by TITA last updated on 20/Dec/20
please help
$$\mathrm{please}\:\mathrm{help} \\ $$
Answered by mr W last updated on 20/Dec/20
let log 2=x, log 3=y log_6 15=((log 15)/(log 6))=((log 3+log 10−log 2)/(log 3+log 2)) =((log 3+1−log 2)/(log 3+log 2))=((y+1−x)/(y+x)) =a ⇒(1+a)x−(1−a)y=1 ...(i) log_(12) 18=((log 18)/(log 12))=((log 2+2 log 3)/(log 3+2 log 2)) =((x+2y)/(y+2x))=b ⇒(1−2b)x+(2−b)y=0 ...(ii) from (i) and (ii) we get x=((b−2)/((1+a)(b−2)−(1−a)(1−2b))) y=((1−2b)/((1+a)(b−2)−(1−a)(1−2b))) log_(25) 24=((log 24)/(log 25))=((log 3+3 log 2)/(2−2 log 2))=((3x+y)/(2(1−x))) =((3(b−2)+(1−2b))/(2[(1+a)(b−2)−(1−a)(1−2b)−b+2])) ⇒((b−5)/(2(2b−a−ab−1)))
$${let}\:\mathrm{log}\:\mathrm{2}={x},\:\mathrm{log}\:\mathrm{3}={y} \\ $$$$\mathrm{log}_{\mathrm{6}} \:\mathrm{15}=\frac{\mathrm{log}\:\mathrm{15}}{\mathrm{log}\:\mathrm{6}}=\frac{\mathrm{log}\:\mathrm{3}+\mathrm{log}\:\mathrm{10}−\mathrm{log}\:\mathrm{2}}{\mathrm{log}\:\mathrm{3}+\mathrm{log}\:\mathrm{2}} \\ $$$$=\frac{\mathrm{log}\:\mathrm{3}+\mathrm{1}−\mathrm{log}\:\mathrm{2}}{\mathrm{log}\:\mathrm{3}+\mathrm{log}\:\mathrm{2}}=\frac{{y}+\mathrm{1}−{x}}{{y}+{x}}\:={a} \\ $$$$\Rightarrow\left(\mathrm{1}+{a}\right){x}−\left(\mathrm{1}−{a}\right){y}=\mathrm{1}\:\:\:…\left({i}\right) \\ $$$$\mathrm{log}_{\mathrm{12}} \:\mathrm{18}=\frac{\mathrm{log}\:\mathrm{18}}{\mathrm{log}\:\mathrm{12}}=\frac{\mathrm{log}\:\mathrm{2}+\mathrm{2}\:\mathrm{log}\:\mathrm{3}}{\mathrm{log}\:\mathrm{3}+\mathrm{2}\:\mathrm{log}\:\mathrm{2}} \\ $$$$=\frac{{x}+\mathrm{2}{y}}{{y}+\mathrm{2}{x}}={b} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{2}{b}\right){x}+\left(\mathrm{2}−{b}\right){y}=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get} \\ $$$${x}=\frac{{b}−\mathrm{2}}{\left(\mathrm{1}+{a}\right)\left({b}−\mathrm{2}\right)−\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−\mathrm{2}{b}\right)} \\ $$$${y}=\frac{\mathrm{1}−\mathrm{2}{b}}{\left(\mathrm{1}+{a}\right)\left({b}−\mathrm{2}\right)−\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−\mathrm{2}{b}\right)} \\ $$$$ \\ $$$$\mathrm{log}_{\mathrm{25}} \:\mathrm{24}=\frac{\mathrm{log}\:\mathrm{24}}{\mathrm{log}\:\mathrm{25}}=\frac{\mathrm{log}\:\mathrm{3}+\mathrm{3}\:\mathrm{log}\:\mathrm{2}}{\mathrm{2}−\mathrm{2}\:\mathrm{log}\:\mathrm{2}}=\frac{\mathrm{3}{x}+{y}}{\mathrm{2}\left(\mathrm{1}−{x}\right)} \\ $$$$=\frac{\mathrm{3}\left({b}−\mathrm{2}\right)+\left(\mathrm{1}−\mathrm{2}{b}\right)}{\mathrm{2}\left[\left(\mathrm{1}+{a}\right)\left({b}−\mathrm{2}\right)−\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−\mathrm{2}{b}\right)−{b}+\mathrm{2}\right]} \\ $$$$\Rightarrow\frac{{b}−\mathrm{5}}{\mathrm{2}\left(\mathrm{2}{b}−{a}−{ab}−\mathrm{1}\right)} \\ $$
Commented by Dwaipayan Shikari last updated on 20/Dec/20
We have 9° C in Kolkata
$${We}\:{have}\:\mathrm{9}°\:{C}\:{in}\:{Kolkata} \\ $$
Commented by MJS_new last updated on 20/Dec/20
lol I′m baking cookies so I′m not as fast as you...
$$\mathrm{lol}\:\mathrm{I}'\mathrm{m}\:\mathrm{baking}\:\mathrm{cookies}\:\mathrm{so}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{as}\:\mathrm{fast}\:\mathrm{as}\:\mathrm{you}… \\ $$
Commented by mr W last updated on 20/Dec/20
i′m just eating cookies with Glu^ hwein, so i must be faster...
$${i}'{m}\:{just}\:{eating}\:{cookies}\:{with}\:{Gl}\ddot {{u}hwein}, \\ $$$${so}\:{i}\:{must}\:{be}\:{faster}… \\ $$
Commented by MJS_new last updated on 20/Dec/20
I made Elisenlebkuchen, Vanillekipferl & Zimtsterne... season greetings from Vienna!
$$\mathrm{I}\:\mathrm{made}\:\mathrm{Elisenlebkuchen},\:\mathrm{Vanillekipferl}\:\& \\ $$$$\mathrm{Zimtsterne}…\:\mathrm{season}\:\mathrm{greetings}\:\mathrm{from}\:\mathrm{Vienna}! \\ $$
Commented by Dwaipayan Shikari last updated on 20/Dec/20
What is the weather in Vienna sir? May be cold
$${What}\:{is}\:{the}\:{weather}\:{in}\:{Vienna}\:{sir}?\:{May}\:{be}\:\:{cold}\: \\ $$
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
it′s now 5:40 p.m., sunset was 4:02 p.m. we have +3°C and mist / light rain
$$\mathrm{it}'\mathrm{s}\:\mathrm{now}\:\mathrm{5}:\mathrm{40}\:\mathrm{p}.\mathrm{m}.,\:\mathrm{sunset}\:\mathrm{was}\:\mathrm{4}:\mathrm{02}\:\mathrm{p}.\mathrm{m}. \\ $$$$\mathrm{we}\:\mathrm{have}\:+\mathrm{3}°\mathrm{C}\:\mathrm{and}\:\mathrm{mist}\:/\:\mathrm{light}\:\mathrm{rain} \\ $$
Commented by talminator2856791 last updated on 22/Dec/20
in south africa it is around 20° in morning now
$$\:\mathrm{in}\:\mathrm{south}\:\mathrm{africa}\:\mathrm{it}\:\mathrm{is}\:\mathrm{around}\:\mathrm{20}°\:\mathrm{in}\:\mathrm{morning}\:\mathrm{now} \\ $$
Answered by MJS_new last updated on 20/Dec/20
a=((ln 15)/(ln 6))=((ln 3 +ln 5)/(ln 2 +ln3)) b=((ln 18)/(ln 12))=((ln 2 +2ln3)/(2ln 2 +ln 3)) let ln 2 =p; ln 3 =q; ln 5 =r a=((q+r)/(p+q)) ⇒ p=((−qa+q+r)/a) b=((p+2q)/(2p+q))=((−qa−q−r)/(qa−2q−2r)) ⇒ q=(((2b−1)r)/(ab+a−2b+1)) ⇒ p=(((2−b)r)/(ab+a−2b+1)) we need to know ((ln 24)/(ln 25))=((3ln 2 +ln 2)/(2ln 5))=((3p+q)/(2r)) and inserting we get ((5−b)/(2(ab+a−2b+1)))
$${a}=\frac{\mathrm{ln}\:\mathrm{15}}{\mathrm{ln}\:\mathrm{6}}=\frac{\mathrm{ln}\:\mathrm{3}\:+\mathrm{ln}\:\mathrm{5}}{\mathrm{ln}\:\mathrm{2}\:+\mathrm{ln3}} \\ $$$${b}=\frac{\mathrm{ln}\:\mathrm{18}}{\mathrm{ln}\:\mathrm{12}}=\frac{\mathrm{ln}\:\mathrm{2}\:+\mathrm{2ln3}}{\mathrm{2ln}\:\mathrm{2}\:+\mathrm{ln}\:\mathrm{3}} \\ $$$$\mathrm{let}\:\mathrm{ln}\:\mathrm{2}\:={p};\:\mathrm{ln}\:\mathrm{3}\:={q};\:\mathrm{ln}\:\mathrm{5}\:={r} \\ $$$${a}=\frac{{q}+{r}}{{p}+{q}}\:\Rightarrow\:{p}=\frac{−{qa}+{q}+{r}}{{a}} \\ $$$${b}=\frac{{p}+\mathrm{2}{q}}{\mathrm{2}{p}+{q}}=\frac{−{qa}−{q}−{r}}{{qa}−\mathrm{2}{q}−\mathrm{2}{r}}\:\Rightarrow\:{q}=\frac{\left(\mathrm{2}{b}−\mathrm{1}\right){r}}{{ab}+{a}−\mathrm{2}{b}+\mathrm{1}} \\ $$$$\Rightarrow\:{p}=\frac{\left(\mathrm{2}−{b}\right){r}}{{ab}+{a}−\mathrm{2}{b}+\mathrm{1}} \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{know}\:\frac{\mathrm{ln}\:\mathrm{24}}{\mathrm{ln}\:\mathrm{25}}=\frac{\mathrm{3ln}\:\mathrm{2}\:+\mathrm{ln}\:\mathrm{2}}{\mathrm{2ln}\:\mathrm{5}}=\frac{\mathrm{3}{p}+{q}}{\mathrm{2}{r}} \\ $$$$\mathrm{and}\:\mathrm{inserting}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{5}−{b}}{\mathrm{2}\left({ab}+{a}−\mathrm{2}{b}+\mathrm{1}\right)} \\ $$
Answered by Dwaipayan Shikari last updated on 20/Dec/20
((log15)/(log6))=a ⇒log(5)+log(3)=alog(3)+alog(2)→(a) ((log(18))/(log(12)))=b⇒2log(3)+log(2)=blog(3)+2blog(2)⇒((log3)/(log2))=((2b−1)/(2−b)) From ( a ) log(5)=(a−1)log3+alog(2)=log(2)((((a−1)(2b−1))/(2−b))+a) ((log(8)+log(3))/(2log(5)))=((3log(2)+log(2)(((2b−1)/(2−b))))/(2log(2)(((2ab−2b−a+1+2a−ab)/(2−b)))))=((5−b)/(2(ab+a−2b+1)))
$$\frac{{log}\mathrm{15}}{{log}\mathrm{6}}={a}\:\:\:\Rightarrow{log}\left(\mathrm{5}\right)+{log}\left(\mathrm{3}\right)={alog}\left(\mathrm{3}\right)+{alog}\left(\mathrm{2}\right)\rightarrow\left({a}\right) \\ $$$$\frac{{log}\left(\mathrm{18}\right)}{{log}\left(\mathrm{12}\right)}={b}\Rightarrow\mathrm{2}{log}\left(\mathrm{3}\right)+{log}\left(\mathrm{2}\right)={blog}\left(\mathrm{3}\right)+\mathrm{2}{blog}\left(\mathrm{2}\right)\Rightarrow\frac{{log}\mathrm{3}}{{log}\mathrm{2}}=\frac{\mathrm{2}{b}−\mathrm{1}}{\mathrm{2}−{b}} \\ $$$${From}\:\left(\:{a}\:\right)\:\:\:{log}\left(\mathrm{5}\right)=\left({a}−\mathrm{1}\right){log}\mathrm{3}+{alog}\left(\mathrm{2}\right)={log}\left(\mathrm{2}\right)\left(\frac{\left({a}−\mathrm{1}\right)\left(\mathrm{2}{b}−\mathrm{1}\right)}{\mathrm{2}−{b}}+{a}\right) \\ $$$$\frac{{log}\left(\mathrm{8}\right)+{log}\left(\mathrm{3}\right)}{\mathrm{2}{log}\left(\mathrm{5}\right)}=\frac{\mathrm{3}{log}\left(\mathrm{2}\right)+{log}\left(\mathrm{2}\right)\left(\frac{\mathrm{2}{b}−\mathrm{1}}{\mathrm{2}−{b}}\right)}{\mathrm{2}{log}\left(\mathrm{2}\right)\left(\frac{\mathrm{2}{ab}−\mathrm{2}{b}−{a}+\mathrm{1}+\mathrm{2}{a}−{ab}}{\mathrm{2}−{b}}\right)}=\frac{\mathrm{5}−{b}}{\mathrm{2}\left({ab}+{a}−\mathrm{2}{b}+\mathrm{1}\right)} \\ $$
Commented by Dwaipayan Shikari last updated on 20/Dec/20
Slowest possible answer 😁
$${Slowest}\:{possible}\:{answer}\: \\ $$😁

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *