# Solve exponential equation

#### Virgin Vanilla

##### New member

$$\displaystyle 3^{2x}-4.(3^{x})+3=0$$

Since the base is same;

$$\displaystyle 2x+x+1=1\\ 3x=1-1\\ x=0/3\\ x=0$$

I took base as 3 regardless of 4. Is it correct?

#### Subhotosh Khan

##### Super Moderator
Staff member

$$\displaystyle 3^{2x}-4.(3^{x})+3=0$$

Since the base is same;

$$\displaystyle 2x+x+1=1\\ 3x=1-1\\ x=0/3\\ x=0$$

I took base as 3 regardless of 4. Is it correct?
$$\displaystyle 3^{2x}-4.(3^{x})+3=0$$

$$\displaystyle (3^{x})^2-4.(3^{x})+3=0$$

Substitute:

u = 3x

we have now a Quadratic Equation

u2 - 4*u + 3 = 0

continue....

#### Virgin Vanilla

##### New member
$$\displaystyle u2 - 4*u + 3 = 0$$
$$\displaystyle u-3=0$$ or $$\displaystyle u-1=0$$
$$\displaystyle u=3$$ or $$\displaystyle u=1\\\\$$
if $$\displaystyle u=3$$
$$\displaystyle 3^x=3$$
$$\displaystyle x=1\\\\$$
if $$\displaystyle u=1$$
$$\displaystyle 3^x=1\\ x=1$$

#### HallsofIvy

##### Elite Member
-

$$\displaystyle 3^{2x}-4.(3^{x})+3=0$$

Since the base is same;

$$\displaystyle 2x+x+1=1$$
No, this step is not valid. Instead, let $$\displaystyle y= 3^x$$ so the equation becomes
$$\displaystyle y^2- 4y+ 3= (y- 3)(y- 1)= 0$$.

$$\displaystyle y= 3^x= 3$$ which means x= 1. This is what you had before.
$$\displaystyle y= 3^x= 1$$. Before, and this is what Harry-the Cat called to your attention, you had x= 1 which is not correct. Any number to the 0 power is 1 so $$\displaystyle 3^x= 1$$ leads to x= 0.
This is simply w
$$\displaystyle 3x=1-1\\ x=0/3\\ x=0$$

I took base as 3 regardless of 4. Is it correct?
I'm not sure why you say that. Every x is an exponent of 3 so of course the base is 3. In any case, this equation has two roots, x= 0 and x= 1.