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Question Number 50932 by Tawa1 last updated on 22/Dec/18

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Dec/18

$${log}_{{\left({log}_{a}c\right)}^2}{log}_{b}a=\frac{-3}{2}$$$${\left[{\left({log}_{a}c\right)}^2\right]}^{\frac{-3}{2}}={log}_{b}a$$$${\left({log}_{a}c\right)}^{-3}={log}_{b}a$$$$let{log}_{e}a=lna=p$$$$lnb=qlnc=r$$$${\left(\frac{lnc}{lna}\right)}^{-3}=\left(\frac{lna}{lnb}\right){\left(\frac{r}{p}\right)}^{-3}=\left(\frac{p}{q}\right)$$$$\frac{1}{{\left(\frac{r}{p}\right)}^3}=\frac{p}{q}\frac{p}{q}\times \frac{r^3}{p^3}=1p^2=\frac{r^3}{q}....\left(1\right)eqn$$$${log}_{c}b-{log}_{a}c=2$$$$\frac{q}{r}-\frac{r}{p}=2$$$$pq-r^2=2rp....\left(2\right)eqn$$$$wait...$$$${log}_a\left(b^{2018}{c}^{2019}\right)=x$$$$a^x=b^{2018}{c}^{2019}$$$$xlna=2018lnb+2019lnc$$$$x=\frac{2018q+2019r}{p}$$$$x=\frac{2000q+2000r}{p}+\frac{18q+19r}{p}$$$$x=1000\left(\frac{2q+2r}{p}\right)+\left(\frac{18q+19r}{p}\right)$$$$soremainder...$$$$\left(\frac{18q+19r}{p}\right)$$$$=18\left(\frac{q}{p}\right)+19\left(\frac{r}{p}\right)$$$$=18\left(\frac{lnb}{lna}\right)+19\left(\frac{lnc}{lna}\right)$$$$=18{log}_{a}b+19{log}_{a}c$$$$plscheck...$$

Commented by Tawa1 last updated on 24/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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