Solve the equation: 3^( 3x^2 + 5x - 4) = 9^-3

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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3^(3x^2 + 5x - 4) = 9^-3

First we will simplify:

we know that 9 = 3^2

==> 3^(3x^2 + 5x - 4) = (3^2)^-3

But we ww know that:  (a^b)^c = a^(b*c)

==> 3^(3x^2 + 5x -4) = 3^-6

==> Now since the bases are equals (3) , then the powers are equal too.

==> 3x^2 + 5x -4 = -6

==> 3x^2 + 5x + 2 = 0

Now we will solve :

x1= [-5 + sqrt(25- 4*3*2)]/ 6

    = (-5 + 1)/6 = -4/6 = -2/3

==> x1= -2/3

==> x2= (-5-1)/6 = -6/6 = -1

==> x2= -1

Then the answer is:

x= { -2/3 , -1}

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve 3^ (3x^2 + 5x - 4) = 9^-3 for x.

First we convert both the sides as terms to the power 3

Therefore 3^ (3x^2 + 5x - 4) = 9^-3

=> 3^ (3x^2 + 5x - 4) = (3^2) ^-3

=> 3^ (3x^2 + 5x - 4) = 3^-6

Now take the logarithm to the base 3 of both the sides. We can do this as both the sides are terms raised to the power 3 and 3 being positive the logarithm is defined.

=> log 3[3^ (3x^2 + 5x - 4)] = log 3(3^-6)

Now, use the relation: log a (a^b) = b

=> 3x^2 + 5x - 4 = -6

=> 3x^2 + 5x - 4 + 6 = 0

=> 3x^2 + 5x + 2 = 0

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

=> 3x(x+1) + 2(x+1) =0

=> (3x+2) (x+1) =0

For 3x+2 =0 we have x= -2/3

and for x+1 =0, we have x = -1

Therefore the required values of x are -2/3 and -1

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve for x. 3^(3x^2+5x-4) = 9^(-3)

We express  the right side  as the exponent of 3:

9^(-3) = {3^(2}^(-3) = 3^(-6).

So now we can rewite the equation as:

3^(3x^2+5x -4) = 3^-6).

3x^2 + 5x -4 = -6, as a^b = a^c imp[lies b= x.

3x^2 +5x-4+6 = 0.

3x^2 +5x +2 = 0.

3x ^2+3x+2x +2 = 0.

3x(x+1)+2(x+1) = 0.

(x+1)(3x+2) = 0.

 x+1 = 0 or 3x+2 = 0.

 x= -1 or  x= -2/3.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we notice that the right side could be written as a power of 3.

9 = 3^2

We'll also use the rule of negative power:

a^-b = 1/a^b

We'll put a = 9 and b = -3

9^-3 = 1/9^3 = 1/(3^2)^3 = 1/3^6 = 3^-6

We'll re-write the equation:

3^(3x^2 + 5x - 4) = 3^-6

Since the bases are matching, we'll apply one to one rule and we'll put in relation of equality the exponents of both sides:

3x^2 + 5x - 4 = -6

3x^2 + 5x - 4+6 = 0

3x^2 + 5x + 2 = 0

We'll apply the quadratic formula:

 x1 = [-5+sqrt(25 - 24)]/6

x1 = (-5+1)/6

x1 = -2/3

x2 = -1

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