# what is simpler way to solve simultaneos equations log base 3 (xy) = 1 + log base 3 (x+y) and 1/x-1/y=9^-1?

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You may convert 1 into a logarithm of base 3 such that:

`log_3(xy) = log_3 (3) + log_3 (x+y)`

You should convert the sum of logarithms to the right side into the logarithm of the product such that:

`log_3(xy) = log_3 (3(x+y)) `

Since the base of logarithms are the same, hence, you may equate the numbers such that:

`(xy) = 3(x+y)`

Hence, you should consider as the first equation of the system, `(xy) = 3(x+y).`

You need to focus on the second equation of the system and you should bring the terms to a common denominator such that:

`(y - x)/xy = 1/9 =gt -9x + 9y = xy`

You may substitute 3(x+y) for xy in the second equation such that:

`-9x + 9y = 3x + 3y =gt -12x + 6y = 0`

You should divide by -6 both sides such that:

`2x - y = 0 =gt 2x = y`

You may substitute 2x for y in the first equation such that:

`(2x^2) = 3(x+2x) =gt 2x^2 - 9x = 0`

Factoring out x yields:

`x(2x - 9) = 0`

`x_1 = 0 =gt y_1 = 0`

`2x - 9 = 0 =gt 2x = 9 =gt x_2 = 9/2 =gt y_2 = 9`

Notice that it is impossible for x and y to be 0 since `log_3 0 ` does not exist.

**Hence, the solution to the system of equations is `x = 9/2 ; y = 9.` **