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I have the following equation and I'm supposed to choose whether or not it's true.

$\log_7(5x^2) = 2 \log_7(5x)$

According to my textbook, the solution is False because it should look like this instead:

$\log_7(5x^2) = 2 \log_7(\sqrt5x)$

I don't really get how the square root got in there. Shouldn't the first equation be true since the square is only on the $x$ and, therefore, it would become:

$\log_7 5 + 2 \log_7 x $

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    $\begingroup$ Your book is wrong! It should be $2\log_7(\sqrt{5}|x|)$ ! $\endgroup$ – Jakobian Jun 1 '18 at 21:48
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    $\begingroup$ " Shouldn't the first equation be true since the square is only on the x and, therefore" ... No. That is precisely why the first equation is false. For the first equation to be true, the square would have to be on the $(5x)^2$ not $5(x^2)$. $\endgroup$ – fleablood Jun 2 '18 at 1:44
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$\log_7 (5x^2) \ne 2 \log_7 (5x)$ because $x$ is being squared, not $(5x)$.

In actuality: $\log_7(5x^2) = \log_7 5 + \log_7 x^2 = \log_7 5 + 2\log_7 x$

[Assuming $x > 0$. Otherwise we should assume $\log_7 x^2 = \log_7 (-x)^2 = 2\log_7 |x|$].

Whereas had the question been instead $\log_7 (5x)^2$ then $\log_7 (5x)^2 = 2\log_7 5x$.

In answering the question, the text is assuming you, the student, will make an error and say "Well, gosh.... Isn't $\log_7(5x^2) = \log_7 (5x)^2 = 2 \log_7 5x$?" (Presumably right before you eat dirt and shove a crayon up your nose).

In which case the book is shaking its head sadly and the sorry excuse it presumes its students must be, and explains that, no $5x^2=5*(x)^2 \ne (5x)^2 $ but $5x^2 = ([\sqrt 5] x)^2$ so $\log_7(5x^2)$ would actually by $\log_7(5x^2) = \log_7([\sqrt 5]x)^2 = 2\log_7([\sqrt 5] x)$ (And as Adam points out in a comment you book would still be wrong if we don't know $x > 0$).

So that you didn't see where the $\sqrt 5$ came from just means you aren't as dumb as the book assumes you to be.

Note: $\log_7 5 + 2\log_7 |x| = 2\log_7 \sqrt 5|x|$ (and $\log_7 5 = \log_7 \sqrt 5^2 = 2\log_7 \sqrt 5$) but the RHS is certain less natural and more convoluted than the LHS.

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  • $\begingroup$ Thank you, this is exactly the explanation I was looking for. Just wish I didn't feel like sticking a crayon up my nose now... $\endgroup$ – blizz Jun 2 '18 at 13:58
  • $\begingroup$ It kind of bothers me when a book assumes a student will make an error. Otherwise deciding $\log 5x^2 = 2\log \sqrt{5}|x|$ is, although correct, preposterously convoluted. $\endgroup$ – fleablood Jun 2 '18 at 16:03
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Since $\log_a N^b = b\log_a N$ you have $$2\log_7 (\sqrt{5}x) = \log_7 (\sqrt{5}x)^2 = \log_7 (5x^2).$$ Another property of logarithms is that $\log ab = \log a + \log b$. So $$\log_7 (5x^2) = \log_7 5 + \log_7 x^2 = \log_7 5 + 2\log_7 \lvert x\rvert.$$

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  • $\begingroup$ But would this apply if the exponent is directly on the x, not outside the parentheses? $\endgroup$ – blizz Jun 1 '18 at 21:47
  • $\begingroup$ I edited the answer. Those steps show that you cannot take the 2 out if it is only exponent of $x$. $\endgroup$ – Gibbs Jun 1 '18 at 21:47
  • $\begingroup$ Based on your example, wouldn't it mean that the original equation should result in log7 of 25x^2? $\endgroup$ – blizz Jun 1 '18 at 21:55
  • $\begingroup$ @blizz No. Why? $\endgroup$ – Gibbs Jun 1 '18 at 21:56
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$$\log_7(5x^2) = \log_7(\sqrt 5|x|)^2 = 2\log_7(\sqrt 5|x|)$$

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