To solve the equation 9^x-4*3^x+3 = 0.

Soltion:

9^x = (3^x)^2

So Ppt 3^x = y, then the equation becomes:

y^2 -4y+3 = 0 which is quadratic in y. So it couls be solved in the uasual way a quadratic equation is solved. We factorise the left of y^2-4y+3 = 0. So,

(y-3)(y-1) =0. Or

y-3 = gives y = 3. Or 3^x = 3 = 3^1 Or x = 1.

y-1 = 0 gives y = 1. Or 3^x = 1= 3^0. So x = 0

The equation is an exponential equation and we could write it in this way:

3^2x + 4*3^x + 3 = 0

It is advisable to solve this type of equation, using substitution method.

Now we can do the substitution 3^x=t

t^2 + 4*t + 3 = 0

This is a quadratic equation and to find it's roots, we can apply the quadratic formula:

t1 = [-4+sqrt(16-12)]/2

t1 = (-4+2)/2

t1 = -1

t2 = (-4-2)/2

t2 = -3

But the initial equation is not solved yet.

3^x = t1

3^x = -1, impossible because 3^x>0!

3^x = t2

3^x = -3, again impossible, because 3^x>0!

The equation has not real solutions!

The equation 9^x -4*3^x + 3 = 0 has to be solved.

Notice that 9^x can be written as (3^2)^x. Use the property (a^b)^c = (a^c)^b

This gives:

(3^x)^2 - 4*3^x + 3 = 0

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

3^x(3^x - 3) - 1(3^x - 3) = 0

(3^x - 3)(3^x - 1) = 0

3^x = 3 and 3^x = 1

x = 1 and x = 0

The solution of the equation is x = 1 and x = 0