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)`.

Explanation

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`.