Equations with logarithms and unknown values So, when doing some problems involving logarithms I noticed that the implications didn't seem to be going one way or another. So the step from:
$\log_{10}x^2$ = 6
to
$\log_{10}x$ = 3
isn't necessarily true espescially for a negative value of $x$. This seems to suggest that we can't really determine whether the implication will necessarily go => or <= unless we know the exact logarithmic equations. This seems to me to be very troubling since in lots of problems we do lots and lots of operations on the logarithms.
So does this mean that the answers we get after doing lots of operations on our logarithms invalid (or very difficult to determine whether they are or not)? Surely, this can't be the case and I'm missing something here. Could somebody please explain? Hopefully this was clear... Thanks in advance!
 A: 
In the above Venn diagrams, element $x_1$ of set $A$ represents solution $x_1$ of equation $A(x).$ So:

*

*$A(x){\implies}B(x):\;$ whichever $x$ satisfies equation
$A(x)$ also satisfies equation $B(x)$

*$A(x){\kern.6em\not\kern-.6em\implies}B(x):\;$ some $x$ satisfies equation $A(x)$ but not equation $B(x)$

*$A(x){\iff}B(x):\;$ equations $A(x)$ and $B(x)$ have the
same solution set.
\begin{gather}\lg x = 3 \implies \lg x^2 = 6\implies x=\pm1000\tag1\\
\lg x^2 = 6 \kern.6em\not\kern-.6em\implies \lg x = 3\implies x=1000\tag2\\
x^2 = 25 \kern.6em\not\kern-.6em\implies x = 5\tag3\\
x^2=25\implies |x|^2=25\implies |x|=5\implies x=\pm5\tag4\\
\lg x^2 = 6 \implies \lg |x|^2 = 6\implies \lg |x| = 3\implies x=\pm1000\tag5\end{gather}
Of the eleven operations above, only the third and fifth ones discard solution(s), so they are invalid steps.
The other nine operations do not discard any required solution, and are valid steps. However, valid operations may nevertheless create extraneous solution(s), as in $(1)$ above. I wrote here about how to avoid extraneous solutions (it's got to do with $‘\iff$’), but it is safer (less carelessness-prone) to just treat obtained solutions as merely candidate (provisional) solutions, then sift out those that fail to satisfy the given equation. Thus, extraneous solutions are not really a problem.
A: No, all the answers should be valid if the inputs $x$ are in the domain of the log function and are positive real numbers with regard to the curriculum you are studying. If $x > 0$ then your implication is valid. When reading intermediate algebra books, you notice that the assumption is that $x > 0$ for the properties to hold. This means that whenever dealing with logarithm $\log_a x$, you must take into consideration at the start that $x > 0$.
A: The only thing invalid in your posting is your presumption that $\sqrt{x^2} = x.$  Your posting becomes valid if you change the equation to $\sqrt{x^2} = |x|.$
That is:
$$\log_{10}x^2 = 6 \iff \log_{10}|x| = 3.$$
