How would solve equation 2^x+7^x-9^x=0?

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You should divide by `9^x` the equation, such that:

`(2/9)^x + (7/9)^x - 1 = 0`

You need to isolate the terms that contain the exponent `x` to the left side, such that:

`(2/9)^x + (7/9)^x = 1`

You need to notice that the associated function `f(x) = (2/9)^x + (7/9)^x` strictly decreases, hence, `f(x) = (2/9)^x + (7/9)^x ` represents an injective function, having an unique solution.

You should notice that for `x = 1` , the equation `(2/9)^x + (7/9)^x = 1` holds, such that:

`(2/9)^1 + (7/9)^1 = 1 => (2+7)/9 = 1 => 9/9 = 1 => 1 = 1`

**Hence, evaluating the solution to the given equation, yields **`x = 1.`

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