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Question Number 192918 by MATHEMATICSAM last updated on 31/May/23

Answered by aba last updated on 31/May/23

$$\mathrm{x}+1={\log }_{2\mathrm{a}}\left(\frac{\mathrm{bcd}}{2}\right)+1={\log }_{2\mathrm{a}}\left(\frac{\mathrm{bcd}}{2}\right)+{\log }_{2\mathrm{a}}\left(2\mathrm{a}\right)={\log }_{2\mathrm{a}}\left(\mathrm{abcd}\right)\Rightarrow \frac{1}{\mathrm{x}+1}={\log }_{\mathrm{abcd}}\left(2\mathrm{a}\right)$$$$\mathrm{y}+1={\log }_{3\mathrm{b}}\left(\frac{\mathrm{acd}}{3}\right)+1={\log }_{3\mathrm{b}}\left(\mathrm{abcd}\right)\Rightarrow \frac{1}{\mathrm{y}+1}={\log }_{\mathrm{abcd}}\left(3\mathrm{b}\right)$$$$\mathrm{z}+1={\log }_{4\mathrm{c}}\left(\frac{\mathrm{abd}}{4}\right)+1={\log }_{4\mathrm{c}}\left(\mathrm{abcd}\right)\Rightarrow \frac{1}{\mathrm{z}+1}={\log }_{\mathrm{abcd}}\left(4\mathrm{c}\right)$$$$\mathrm{p}+1={\log }_{5\mathrm{d}}\left(\frac{\mathrm{abc}}{5}\right)+1={\log }_{5\mathrm{d}}\left(\mathrm{abcd}\right)\Rightarrow \frac{1}{\mathrm{p}+1}={\log }_{\mathrm{abcd}}\left(5\mathrm{d}\right)$$$$\frac{1}{\mathrm{x}+1}+\frac{1}{\mathrm{y}+1}+\frac{1}{\mathrm{z}+1}+\frac{1}{\mathrm{p}+1}={\log }_{\mathrm{abcd}}(5!\mathrm{abcd})={\log }_{\mathrm{abcd}}(5!)+1$$$$\Rightarrow {\log }_{\mathrm{abcd}}(5!)+1={\log }_{\mathrm{abcd}}\left(\mathrm{N}\right)+1\Rightarrow \mathrm{N}=5!=120$$

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