Finding $x$ using logs - two methods give different answers

Question

I have been presented with the following question:

$$\log_3x +\log_x3 = 2.$$

I convert $\log_x3$ to base $3$ giving:

$$\log_3x +\frac{1}{\log_3x} = 2.$$

Then allow $\log_3x = u$ giving :

$$u+\frac{1}{u}=2.$$

I then multiply by $u$ and subtract two to give

$$u^2-2u + 1 = 0.$$

From here I find $u$ using two methods.

Method $1$ - using the quadratic formula which gives me $u=1.$

Method 2 - finding 2 numbers (factors) that adds to $-2$ and multiplies to $1$, this gives me $u=-1.$

Of course I then have two different results for $x$.

Please can someone show me where I have gone wrong?

EDIT - method 2 finds factors not roots, this was my error.

Answer 1

The equation $u^2-2u+1=0$ is equivalent to $$ (u-1)^2 = 0 $$ which has one real solution $u=1$. So $\log_3 x = 1$ and $x=3$.

When you use method 2 (finding two numbers that add to $−2$ and multiply to give $1$) - this is where the error is. The two roots should actually add to $2$! (Recall Vieta's formulas.)

Answer 2

In method 2 you are trying to factor the polynomial. Note that $(-1) + (-1) = -2$ and $(-1)(-1) = 2$ so we have $u^2 - 2u + 1 = (u + (-1))(u + (-1)) = (u - 1)^2$. Thus $(u - 1)^2 = 0$ so $u = 1$, which agrees with the other answer. The reason this factoring method works is because $(x + a)(x + b) = x^2 + (a + b)x + ab$.

Answer 3

The quadratic formula gives a root,

the other method: $u$ and $v$ in $ax^2 +bx +c$ where $u*v=a*c$ and $u+v=b$, gives a factor.

I need the root in order to find x.