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.

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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 $x \neq -3$.

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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 $x=y$.

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