3^(x^2+4x)=(1/9)^2

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We have to solve 3^(x^2+4x)=(1/9)^2 for x.

3^(x^2+4x)=(1/9)^2

=> 3^(x^2+4x)=(1/3^2)^2

=> 3^(x^2+4x)=(3^-2)^2

=> 3^(x^2+4x)= 3^-4

As the base 3 is the same, we can equate the exponent.

=> x^2 + 4x = -4

=> x^2 + 4x + 4 = 0

=> ( x + 2)^2 = 0

=> x = -2

Therefore x = -2

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