Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm stuck with solving this equation,

$$2 \log x = \log 9 $$

This is how far I made it:

\begin{align} \log x &= \log 4,5 \\ x &= ? \end{align}

I'm a beginner at logarithms so I appreciate ways to solve it and not just an answer.

share|cite|improve this question
up vote 2 down vote accepted

If we start with $$2 \log x = \log 9,$$ the first step is to move the 2, but you can't divide it over like that. The rule is that $a \log b = \log b^a$, so we get $2 \log x = \log x^2$.

Now our equations is $$\log x^2 = \log 9.$$ The next step is to use the fact that $\log A = \log B$ means $A = B$. In our case, that means $$x^2 = 9,$$ and you can solve for $x$. Remember to check that the answer makes sense. For instance $\sqrt{9} = \pm 3$, but you can't take the log of a negative, so $x \neq -3$.

share|cite|improve this answer

Note that $\frac{1}{2}\log{9}\ne\log\frac{9}{2}$. Instead $\frac{1}{2}\log{9}=\log{9}^{1/2}=\log{3}$ - constants become powers when you take them inside the $\log$.

Once you're in the situtation where you have the equation $\log{x}=\log{y}$, then taking exponentials of both sides gives you $x=y$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.