Solve for x: 4^(x^2+x)-4096=0.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to determine the value of x for the equation: 4^(x^2+x) - 4096 = 0.

4^(x^2+x) - 4096 = 0

=> 4^(x^2+x) = 4096

=> 4^(x^2+x) = 4096

=> 4^(x^2+x) = 4^6

As the base is equal, equate the exponent

=> x^2 + x = 6

=> x^2 + 3x - 2x - 6 = 0

=> x(x + 3) - 2(x + 3) = 0

=> (x - 2)(x + 3) = 0

=> x = 2 and x = -3

The required values are x = 2 and x = -3

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

[Again, the editing buttons are not working and I cannot bold the final line. I'll answer the question.]

4^(x^2+x)-4096=0

We'll move 4096 to the right side: 4^(x^2+x) = 4096

We'll write 4096 as a power of 4, to create matching bases.

4^(x^2+x) = 4^6

Since the base are matching, we'll apply one to one property:

x^2+ x = 6

We'll subtract 6 both sides: x^2 + x - 6 = 0

We'll apply quadratic formula:

x1 = [-1 + sqrt(1 + 24)]/2 x1 = (-1 + 5)/2 x1 = 2 x2 = (-1-5)/2 x2 = -3

ANSWER: The solutions of the equation are {-3 ; 2}.

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