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Question Number 113803 by Aina Samuel Temidayo last updated on 15/Sep/20

$$\mathrm{If}\mathrm{a}={\log }_{24}12,\mathrm{b}={\log }_{36}24\mathrm{and}$$$$\mathrm{c}={\log }_{48}36,\mathrm{then}1+\mathrm{abc}\mathrm{is}\mathrm{equal}\mathrm{to}$$$$$$$$\left(\mathrm{A}\right)2\mathrm{ab}\left(\mathrm{B}\right)2\mathrm{ac}\left(\mathrm{C}\right)2\mathrm{bc}\left(\mathrm{D}\right)0$$

Answered by bemath last updated on 15/Sep/20

$$a=\frac{\ln 12}{\ln 24}=\frac{\ln 12}{\ln 2+\ln 12},b=\frac{\ln 24}{\ln 36}=\frac{\ln 2+\ln 12}{\ln 3+\ln 12}$$$$c=\frac{\ln 3+\ln 12}{\ln 4+\ln 12}$$$$a\times b\times c=\frac{\ln 12}{\ln 2+\ln 12}\times \frac{\ln 2+\ln 12}{\ln 3+\ln 12}\times \frac{\ln 3+\ln 12}{\ln 4+\ln 12}$$$$=\frac{\ln 12}{\ln 4+\ln 12}=\log {}_{48}\left(12\right)$$$$1+abc=1+\log {}_{48}\left(12\right)$$$$=\log {}_{48}\left(12\times 48\right)$$$$=\frac{\ln 4+2.\ln 12}{\ln 4+\ln 12}$$$$=\frac{2\ln 2+2\ln 12}{\ln 4+\ln 12}=2\left(\frac{\ln 2+\ln 12}{\ln 4+\ln 12}\right)$$$$=2bc$$

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

$$\mathrm{Thanks}.$$

Answered by som(math1967) last updated on 15/Sep/20

$$\mathrm{C})2\mathrm{bc}$$

Answered by $@y@m last updated on 15/Sep/20

$${24}^a=12.....\left(A\right)$$$${36}^b=24.....\left(B\right)$$$${48}^c=36.....\left(C\right)$$$$From\left(C\right),$$$${\left({48}^c\right)}^b={36}^b$$$${48}^{bc}=24\{usimg\left(B\right)$$$${\left({48}^{bc}\right)}^a={24}^a$$$${48}^{abc}=12\{using\left(A\right)$$$${48}^{abc}\times 48=12\times 48$$$${48}^{1+abc}={24}^2$$$${48}^{1+abc}={36}^{2b}$$$${48}^{1+abc}={48}^{2bc}$$$$1+abc=2bc$$$$$$

Commented by Aina Samuel Temidayo last updated on 15/Sep/20

$$\mathrm{Nice}\mathrm{method}.\mathrm{Thanks}.$$

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