Question Number 50932 by Tawa1 last updated on 22/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Dec/18
![log_((log_a c)^2 ) log_b a=((−3)/2) [(log_a c)^2 ]^((−3)/2) =log_b a (log_a c)^(−3) =log_b a let log_e a=lna=p lnb=q lnc=r (((lnc)/(lna)))^(−3) =(((lna)/(lnb))) ((r/p))^(−3) =((p/q)) (1/(((r/p))^3 ))=(p/q) (p/q)×(r^3 /p^3 )=1 p^2 =(r^3 /q)....(1)eqn log_c b−log_a c=2 (q/r)−(r/p)=2 pq−r^2 =2rp....(2)eqn wait... log_a (b^(2018) c^(2019) )=x a^x =b^(2018) c^(2019) xlna=2018lnb+2019lnc x=((2018q+2019r)/p) x=((2000q+2000r)/p)+((18q+19r)/p) x=1000(((2q+2r)/p))+(((18q+19r)/p)) so remainder... (((18q+19r)/p)) =18((q/p))+19((r/p)) =18(((lnb)/(lna)))+19(((lnc)/(lna))) =18log_a b+19log_a c pls check...](https://www.tinkutara.com/question/Q50944.png)
$${log}_{\left({log}_{{a}} {c}\right)^{\mathrm{2}} } {log}_{{b}} {a}=\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$$\left[\left({log}_{{a}} {c}\right)^{\mathrm{2}} \right]^{\frac{−\mathrm{3}}{\mathrm{2}}} ={log}_{{b}} {a} \\ $$$$\left({log}_{{a}} {c}\right)^{−\mathrm{3}} ={log}_{{b}} {a} \\ $$$${let}\:{log}_{{e}} {a}={lna}={p} \\ $$$${lnb}={q}\:\:\:{lnc}={r} \\ $$$$\left(\frac{{lnc}}{{lna}}\right)^{−\mathrm{3}} =\left(\frac{{lna}}{{lnb}}\right)\:\:\:\:\:\:\left(\frac{{r}}{{p}}\right)^{−\mathrm{3}} =\left(\frac{{p}}{{q}}\right) \\ $$$$\frac{\mathrm{1}}{\left(\frac{{r}}{{p}}\right)^{\mathrm{3}} }=\frac{{p}}{{q}}\:\:\:\:\:\:\:\frac{{p}}{{q}}×\frac{{r}^{\mathrm{3}} }{{p}^{\mathrm{3}} }=\mathrm{1}\:\:\:\:{p}^{\mathrm{2}} =\frac{{r}^{\mathrm{3}} }{{q}}….\left(\mathrm{1}\right){eqn} \\ $$$${log}_{{c}} {b}−{log}_{{a}} {c}=\mathrm{2} \\ $$$$\frac{{q}}{{r}}−\frac{{r}}{{p}}=\mathrm{2} \\ $$$${pq}−{r}^{\mathrm{2}} =\mathrm{2}{rp}….\left(\mathrm{2}\right){eqn}\:\:\: \\ $$$${wait}… \\ $$$${log}_{{a}} \left({b}^{\mathrm{2018}} {c}^{\mathrm{2019}} \right)={x} \\ $$$${a}^{{x}} ={b}^{\mathrm{2018}} {c}^{\mathrm{2019}} \\ $$$${xlna}=\mathrm{2018}{lnb}+\mathrm{2019}{lnc} \\ $$$${x}=\frac{\mathrm{2018}{q}+\mathrm{2019}{r}}{{p}} \\ $$$${x}=\frac{\mathrm{2000}{q}+\mathrm{2000}{r}}{{p}}+\frac{\mathrm{18}{q}+\mathrm{19}{r}}{{p}} \\ $$$${x}=\mathrm{1000}\left(\frac{\mathrm{2}{q}+\mathrm{2}{r}}{{p}}\right)+\left(\frac{\mathrm{18}{q}+\mathrm{19}{r}}{{p}}\right) \\ $$$${so}\:{remainder}… \\ $$$$\left(\frac{\mathrm{18}{q}+\mathrm{19}{r}}{{p}}\right) \\ $$$$=\mathrm{18}\left(\frac{{q}}{{p}}\right)+\mathrm{19}\left(\frac{{r}}{{p}}\right) \\ $$$$=\mathrm{18}\left(\frac{{lnb}}{{lna}}\right)+\mathrm{19}\left(\frac{{lnc}}{{lna}}\right) \\ $$$$=\mathrm{18}{log}_{{a}} {b}+\mathrm{19}{log}_{{a}} {c} \\ $$$${pls}\:{check}… \\ $$
Commented by Tawa1 last updated on 24/Dec/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$