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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!

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3 Answers 3

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![![enter image description here

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.

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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 $x > 0$.

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  • $\begingroup$ Thank you! I appreciate your answer. But I'm looking through my textbook and no such specifications seem to be present. Instead they seem to check the final answer by inputting it back to the original equation. Does this validate the answers? I asked this question because I thought it didn't since it's possible to lose a few solutions through these operations. Please let me know what you think! $\endgroup$
    – Nathan Kim
    Apr 4, 2022 at 0:51
  • $\begingroup$ Yes it does. When you answer SAT type or multiple choice algebra questions, you plug "the answer" back to the question and check if it "satisfies" the "equation". If it does, then the number you come up with "is" the solution. And often times, there is only one such number per question. $\endgroup$
    – Wang YeFei
    Apr 4, 2022 at 0:53
  • $\begingroup$ Are you saying that the solutions that result from manipulations of these logarithms can only constitute the (possibly) few solutions and not necessarily all of them? $\endgroup$
    – Nathan Kim
    Apr 4, 2022 at 0:56
  • $\begingroup$ Of course all of them. $\endgroup$
    – Wang YeFei
    Apr 4, 2022 at 1:01
  • $\begingroup$ Assuming there is no restriction with x > 0 (this is the case in my textbook) with the equations I've shown in my initial question, we lose a solution, specifically -$10^3$. So that means that the resulting equation doesn't take into account all values of x that satisfy the original equation. Please correct me where I'm wrong. Sorry, but I'm just kind of confused right now. Heh. $\endgroup$
    – Nathan Kim
    Apr 4, 2022 at 1:09
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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:

$$\log_{10}x^2 = 6 \iff \log_{10}|x| = 3.$$

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