# 3*2^2x + 2*3^2x = 5*6^x. Find x .

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

let us rewrite :

3*2^2x + 2*3^2x = 5*(2*3)^x

3*2^2x + 2&3^2x = 5*2^x * 3^x

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

Let 2^x = a and 3^x = b

==> 3a^2 - 5ab + 2b^2 = 0

Now let us factor:

(3a - 2b)(a-b) = 0

==> 3a-2b = 0

==> 3a = 2b............(1)

==> a-b = 0

==> a= b ............(2)

==> a= b

==> 2^x = 3^x

**==> x= 0**

We notice that 6^x = (2*3)^x

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

We'll subtract 5*2^x*3^x both sides:

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

We'll divide by 3^2x:

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

We'll note (2/3)^x = t

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

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

We'll re-write the equation in 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:

**x = 0**

(2/3)^x = 2/3

**x = 1**

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