# Find xGive an example of exponential equation with different bases, that is solved by substitution.

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We'll ilustrate the idea with the help of the following example:

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

We'll divide by 3^2x to create matching bases that can be replaced by another variable, later on:

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

We'll replace (2/3)^x by t.

We'll 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 create matching bases, writing 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 this type of equation are {0 ; 1}.