If 3^(x^2)/3^4x=1/81 then x=?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve 3^(x^2)/3^4x = 1/81 for x.

3^(x^2)/3^4x = 1/81

=> 3^(x^2) = 3^4x/81

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

=> 3^(x^2)*3^4 = 3^4x

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

As the base is the same equate the exponent

=> x^2 + 4 = 4x

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

=> (x - 2)^2 = 0

=> x = 2

The required value of x = 2

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Since the bases of the exponetials from the left side are matching, we'll apply quotient property and we'll subtract the superscripts, such as:

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

We'll create matching bases both sides, therefore we'll write 81 = 3^4

3^(x^2 - 4x) = 1/3^4

We'll apply the negative power rule:

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

Since the bases of the exponetials are matching, we'll equate the exponents:

x^2 - 4x = -4

We'll add 4 both sides:

x^2 - 4x+ 4 = 0

We'll recognize the perfect square (x - 2)^2:

(x - 2)^2 = 0 => x1 = x2 = 2

The equation has two equal solutions: x1 = x2 = 2.

madmoni06's profile pic

madmoni06 | Student | (Level 1) Honors

Posted on

Good question...

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