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According to the power law:-

$$\log_a (x^k) =k\log_a x $$

So take the following example:-

$$\log_2 ((-2)^2) $$

On solving $\log_2 4=2$

However, if we use the power law, then on simplifying, $2\log_2 (-2)$ is not defined.

So how do I justify this?

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005

2 Answers2

1

The power law is

$$\forall x>0:\log x^k=k\log x.$$

There is no power law for the negatives.

  • Consider the problem- –  Mar 17 '18 at 11:52
  • 2log_3 y -log_3 (y+4) = 2 –  Mar 17 '18 at 11:52
  • Using algebra, the possible solutions are -3,12. However, -3 is not valid on checking. How would you reason this to be excluded? –  Mar 17 '18 at 11:54
  • @user16701 "$-3$ is outside of our domain, hence the solution is only $12$" – ℋolo Mar 17 '18 at 11:59
  • So can I start by stating that y is positive for 2log_3y to be defined, so when I get a possible solution as -3, I can refer back to this and state this is not a possible solution. Is this a logical reasoning? –  Mar 17 '18 at 12:12
  • @user16701: this is a different question. Please stop the chat. –  Mar 17 '18 at 13:05
  • I understand that this is a different question. However, to ask this is a new question would require reference to the original question. And I believe that repetition would be worse. –  Mar 18 '18 at 10:53
  • @user16701: there is no need of a reference to this question. –  Mar 19 '18 at 08:07
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This dicrepancy is due to the fact that

  • $\log x^2$ is defined for $x\neq 0$ but
  • $2\log x$ is defined for $x>0$

then the two expression are equal $\iff x>0$.

What is true $\forall x\neq 0$ is that

  • $\log x^2=2\log |x|$

where we have used that $\sqrt {x^2}=|x|$.

user
  • 154,566
  • So should the law expression be modified? –  Mar 17 '18 at 10:48
  • @user16701 the definition is valid for positive x values and in this case $\log_a (x^k) =k\log_a x$ is always true https://en.wikipedia.org/wiki/Logarithm#Definition – user Mar 17 '18 at 10:52
  • @user16701 as an extension when $x^k$ is positive but $x$ is not the following holds $\log_a (x^k) =k\log_a |x|$ – user Mar 17 '18 at 10:54
  • @gimusi this is true for even $k$, for odd $k$, $\log_a x^k$ is undefined for all $x\le 0$(needless to say, if our codomain is $\Bbb R$) – ℋolo Mar 17 '18 at 11:56
  • @Holo indeed I’ve stated when $x^k$ is positive and if k is odd $x^k$ is positive only for x positive – user Mar 17 '18 at 12:00
  • @gimusi oh, I read it wrong, sorry. You were completely correct – ℋolo Mar 17 '18 at 12:01
  • @Holo You are welcome, thanks for your comment, many times I get wrong thus I appreciate any check and suggestion given in such constructive form! – user Mar 17 '18 at 12:04