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

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll start by using the negative power rule to the right side:

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

Since the bases are matching, we'll apply one to one rule:

x^2 + 4x = -4

We'll add 4 both sideS:

x^2 + 4x + 4 = 0

We'll apply quadratic formula:

x1 = [-4+sqrt(16 - 16)]/2

x1 = (-4+0)/2

x1 = -2

x2 = -2

The solution of the equation is x = -2.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve for x: 3^(x^2+4x) = (1/9)^2.

Solution:

WE first write the right side with base 3.
RHS = (1/9)^2 = (1/3^2)^2 =  (3^-2)^2 = 3^(-4).

So the given equation  now could be written as: 3^(x^2+4x) = 3^(-4).

Since both sides have the matching bases, we equate the exponents:

 x^2+4x = -4.

 x^2 +4x+4 = 0.

(x+2)^2 = 0.

 So x+2 = 0.

Or x= -2.

Therefore x = -2 is the solution of  3^(x^2+4x) = (1/9)^2^2.

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