What is x if 5*6^x-2*3^2x-3*2^2x=0 ?

Asked on by lessoflot

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve for x given that 5*6^x - 2*3^2x - 3*2^2x = 0

Now 5*6^x - 2*3^2x - 3*2^2x = 0

=> 5*2^x*3^x - 2*3^2x - 3*2^2x = 0

Factor out 2^2x

=> 5*3^x/2^x - 2*3^2x/2^2x - 3 = 0

=> 5*(3/2)^x - 2*(3/2)^2x - 3 = 0

let y = (3/2)^x

=> 5t - 2t^2 - 3 = 0

=> 2t^2 - 5t + 6 =0

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

=> t(2t - 3) - 1 (2t - 3) =0

=> ( t- 1)(2t-3) = 0

t is equal to 1 and 3/2.

Now t = (3/2)^x = 1

=> x = 0

t = (3/2)^x = 3/2

=> x = 1

So the required values of x are 0 and 1.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll multiply by -1 and we'll re-write the equation:

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

We remark that 6^x = (2*3)^x

But (2*3)^x = 2^x*3^x

We'll divide by 3^2x all over:

3*(2/3)^2x - 5*(2/3)^x + 2 = 0

We'll note (2/3)^x = t

We'll square raise both sides and we'll get:

(2/3)^2x = t^2

We'll re-write the equation in the new variable t:

3t^2 - 5t + 2 = 0

We'll apply quadratic formula:

t1 = [5+sqrt(25-24)]/6

t1 = (5+1)/6

t1 = 1

t2 = (5-1)/6

t2 = 2/3

Now, we'll put  (2/3)^x = t1:

(2/3)^x = 1

We'll write 1 = (2/3)^0

(2/3)^x = (2/3)^0

Since the bases are matching, we'll apply one to one property of exponentials:

x = 0

(2/3)^x = 2/3

x = 1

The solutions of the equation are {0 ; 1}.

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