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

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lemjay | High School Teacher | (Level 3) Senior Educator

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To start, let `y = log x` .


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


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



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



Then, factor left side.


Set each factor to zero and solve for y.

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

`13y=19`                                    `y=1`


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`


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


                               `x = 10`

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

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