I'm new to logarithms and I am having trouble solving this equation

$$ \left( \log_3 x \right)^2 + \log_3 (x^2) + 1 = 0.$$

How would I solve this? A step-by-step response would be appreciated.

Also, I know how to solve it with assigning $\log_3 (x)$ as $x$ and solving $x^2 + x + 1 = 0$, and getting the answer from there. I am looking to see how I would do it with just logarithms and no quadratics.


  • $\begingroup$ Note that $\log_3(x^2)=2\log_3(x)$. $\endgroup$ – Zev Chonoles Jul 13 '12 at 1:12
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    $\begingroup$ What you think you know, is wrong; you don't get $x^2+x+1$. "just logarithms and no quadratics" --- but it is a quadratic, so I think you are asking the impossible. $\endgroup$ – Gerry Myerson Jul 13 '12 at 1:19

Hint: The second term is equal to $2\log_3(x)$. Let $w=\log_3(x)$. We are looking at the equation $w^2+2w+1=0$.


You should get the answer $x = \frac{1}{3}$.

Here is why. Solving the quadratic equation $w^2 + 2w + 1=0$ gives $w=-1$ as a repeated root. Substituting $w =\log_3(x)$ in $w = -1$ gives $\log_3(x) = -1$

Which implies $3^{\log_3(x)} = 3^{-1} \implies x = \frac{1}{3}$. (we used the inverse function $3^x$ of our function $\log_3 (x)$ ).

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    $\begingroup$ Can you explain why? $\endgroup$ – draks ... Jul 13 '12 at 7:17
  • $\begingroup$ Note that w^2 + 2 w + 1=(w+1)^2=0, this polynomial has a repeated root w=-1. $\endgroup$ – Mhenni Benghorbal Jul 13 '12 at 23:35

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