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We notice that the given equation is a product of two factors. This product is zero if one of the factors is zero.
We'll cancel each factor:
`e^(3x)` + 3 = 0
`e^(3x)` = -3
There is no such value of x for `e^(3x)` to give a negative value, therefore `e^(3x)` + 3 > 0, but never 0.
We'll cancel the next factor:
`e^(3x)` - 1 = 0
`e^(3x)` = 1
We'll create matching bases both sides:
`e^(3x)` = `e^(0)`
Since the bases are matching, we'll equate the exponents:
3x = 0
x = 0
The solution of the equation is x = 0.
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