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Question Number 108240 by mathdave last updated on 15/Aug/20

Answered by john santu last updated on 15/Aug/20

$$\frac{◻JS△}{★}$$$$2^{a+4}=6\Rightarrow 2^{a+4}=2.3$$$$\Rightarrow 2^{a+3}=3...\left(i\right)$$$$3^{2b+1}=3.2^2\Rightarrow 3^{2b}=2^2;3^b=2$$$$then{\left(2^{a+3}\right)}^{2b}=2^2$$$$2^{2ab+6b}=2^2\Rightarrow ab+3b=1$$$$\Rightarrow ab-1=-3b;ab-2=-3b-1$$$$weobtain{3}^{ab-2}=3^{-3b-1}=\frac{1}{{\left(3^b\right)}^3\times 3}$$$$3^{ab-2}=\frac{1}{24}.$$

Answered by 1549442205PVT last updated on 16/Aug/20

$$2^{\mathrm{a}+4}=6\Rightarrow (\mathrm{a}+4)\ln 2=\ln (2.3)=\ln 2+\ln 3$$$$\Rightarrow \mathrm{aln}2=\ln 3-3\ln 2=\ln \frac{3}{8}\Rightarrow \mathrm{a}=\frac{\ln \frac{3}{8}}{\ln 2}\left(1\right)$$$$3^{2\mathrm{b}+1}=12\Rightarrow (2\mathrm{b}+1)\ln 3=\ln (3.4)\iff$$$$2\mathrm{bln}3+\ln 3=\ln 3+\ln 4\Rightarrow 2\mathrm{bln}3=\ln 4$$$$\Rightarrow \mathrm{b}=\frac{\ln 4}{2\ln 3}=\frac{2\ln 2}{2\ln 3}=\frac{\ln 2}{\ln 3}\left(2\right)$$$$\mathrm{From}\left(1\right)\left(2\right)\mathrm{we}\mathrm{get}\mathrm{ab}=\frac{\ln \frac{3}{8}}{\ln 2}\times \frac{\ln 2}{\ln 3}$$$$=\frac{\ln \frac{3}{8}}{\ln 3}={\log }_3\frac{3}{8}$$$$\Rightarrow \mathrm{ab}-2={\log }_3\frac{3}{8}-2={\log }_3\frac{3}{8}-{\log }_3{3}^2$$$$={\log }_3\left(\frac{3}{8}/9\right)={\log }_3\left(\frac{1}{24}\right).\mathrm{Therefore}$$$$3^{\mathrm{ab}-2}=3^{{\log }_3\left(\frac{1}{24}\right)}=\frac{1}{24}(\mathrm{since}{\mathrm{a}}^{{\log }_{\mathrm{a}}\mathrm{N}}=\mathrm{N})$$$$\mathrm{Consequently}\mathrm{we}\mathrm{choose}\mathrm{C}$$

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