# How to solve `2/(7-log x)+9/(11+log x)=13/12` ?

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`2/(7-logx)+9/(11+logx)=13/12`

To start, let `y = log x` .

`2/(7-y)+9/(11+y)=13/12`

To eliminate the denominators, multiply both sides by the LCD which is `12(7-y)(11+y)` .

`12(7-y)(11+y)*[2/(7-y)+9/(11+y)]=13/12*12(7-y)(11+y)`

`24(11+y)+108(7-y) =13(7-y)(11+y)`

`264+24y+756-108y=13(-y^2-4y+77)`

`1020-84y=-13y^2-52y+1001`

Then, express the equation in quadratic form `ax^2+bx+c=0` .

`1020-84y+13y^2+52y-1001=0`

`13y^2-32y+19=0`

Then, factor left side.

`(13y+1085-19)(y-1)=0`

Set each factor to zero and solve for y.

`13y-19=0` and `y-1=0`

`13y=19` `y=1`

`y=19/13`

To solve for x, substitute the value of y to `y = logx` . Then, express it to its equivalent exponential form which is `10^y=x` .

`y=19/13` , `19/13=log x`

`x=10^(19/13)`

`y=1` , `1 = log x`

`x=10^1`

`x = 10`

**Hence, the solutions are `x=10` and `x=10^(19/13)` .**