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Solve the following equaions for x and y `log_(10)x+(1)/(2)log_(10)x+(1)/(4)log_(10)x+"...."=y " and " (1+3+5+"...."+(2y-1))/(4+7+10+"...."+(3y-1))=(20)/(7log_(10)x)`. |
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Answer» From the first equation ` log_(10)x{1+(1)/(2)+(1)/(4)+"…."+oo}=y` `implies log_(10)x{(1)/(1-(1)/(2))}=y` `implies 2log_(10)x=y" " "….(i)"` From the second equation `(1+3+5+"...."+(2y-1))/(4+7+10+"...."+(3y-1))=(20)/(7log_(10)x)` `implies ((y)/(2)(1+2y-1))/((y)/(2)(4+3y+1))=(20)/(7log_(10)x)` `implies (2y)/(3y+5)=(20)/(7log_(10)x)` `implies 7y(2log_(10)x)=60y+100` `implies 7y(y)=60y+100 " " [" from Eq. (i) "]` `implies 7y^(2)-60y-100=0` `:. (y-10)(7y+10)=0` `:. y=10,yne (-10)/(7)` [because y being the number of terms in series `implies y in N`] From Eq. (i), we have `2log_(10)x=10 implies log_(10)x=5` `:.x=10^(5)` Hence, required solution s `x=10(5),y=10`. |
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