# How solve the equation log(x-7/x+33)-log2=-log(27-x)?

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### 1 Answer

You need to use logarithmic identities such that:

`log a + log b = log (a*b)`

`log a - log b = log (a/b)`

Hence, converting the difference `log((x-7)/(x+33))-log2` into a quotient yields:

`log((x-7)/(x+33))-log2 = log((x-7)/(2(x+33)))`

You need to rewrite the equation such that:

`log((x-7)/(2(x+33))) = -log(27-x)`

`log((x-7)/(2(x+33))) + log(27-x) = 0`

`log((x-7)(27-x)/(2(x+33))) = 0 => (x-7)(27-x)/(2(x+33)) = 10^0`

`((x-7)(27-x)/(2(x+33))) = 1 => (x-7)(27-x) = 2(x+33)`

You need to open the brackets such that:

`27x - x^2 - 189 + 7x - 2x - 66 = ` 0

`- x^2 + 32x - 255 = 0`

`x^2 - 32x + 255 = 0`

`x_(1,2) = (32+-sqrt(1024 - 1020))/2 => x_(1,2) = (32+-sqrt4)/2`

`x_(1,2) = (32+-2)/2 => x_1 = 17 ; x_2 = 15`

**Notice that the logarithms are valid for `x in (7,27), ` hence, evaluating the solutions to the given equation under the given conditions, yields `x_1 = 17 ; x_2 = 15` .**

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