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Consider $\log(x-3)^2$ and $2\log(x-3)$. The second expression defines a function that is the same as the branch of the function that is defined by the first expression for $x>3$. That function is not defined for $x<3$. How then it is the case that this is an identity when the two expression actually define different functions?

LearningMath
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  • I think it's just convenient to write it like this though if $p$ is even then the identity is $\log x^p=p\log |x|$ – kingW3 Jun 18 '17 at 20:42

3 Answers3

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There some cases

$1$. If you know that $x>0$ and $p$ is any real number then you can write it just as $$\log { x^{ p } } =p\log x$$ $2$. If you haven't got any information about $x$ but you know $p$ is even number ,then you should write it as $$\\ \log { x^{ p } } =p\log \left| x \right| \\ \\ \\ \\ $$

haqnatural
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  • What about when $p$ is odd? – LearningMath Jun 18 '17 at 20:55
  • it actually includes to the case $1$ inverse version, if p is odd then logarithmic expression should be positive – haqnatural Jun 18 '17 at 21:00
  • @LearningMath When $p$ is an odd integer neither of the sides is definied for $x\le 0$. –  Jun 18 '17 at 21:00
  • When we don't have information about $x$ which means $x$ is some expression then we just take the expression as it is when solving equations and at the end we have to check the solutions we get because the final set of solutions can be larger. And if we have $p$ even, then take the absolute value because this also should be considered (there can be values that make the expression positive too) This is how I understand it now. One more question: Is it possible when solving logarithmic equations to arrive to a solution set that contains only the solutions of the original equation not other ones? – LearningMath Jun 18 '17 at 21:10
  • You understand correctly.Note that the domain of $log$ should be always positive, in another case(when it negative) it makes nonsense.At the end of the solution, you should intersect the set of solutions with the domain of $log$ – haqnatural Jun 18 '17 at 21:17
  • Thank you. I wonder, is there any method to be sure always to arrive to equations that have solutions the same as the original equation? I mean, not having to check them when you are finished solving the equation. – LearningMath Jun 18 '17 at 21:20
  • To check means you put roots in the equation to be sure that they satisfy equation, if you don't want to do this, find domain before solution then observe that if roots includes to this domain or not, hope I could explain – haqnatural Jun 18 '17 at 21:26
  • I suppose also if we have a radical expression we also have to take the absolute value? For example $x^{\frac{2}{3}}$. Am I right? Edit: for example: $\log(x-3)^{\frac{2}{3}} = \frac{2}{3}\log\left | x-3 \right |$ – LearningMath Jun 18 '17 at 21:42
  • @LearningMath, yes you are right – haqnatural Jun 18 '17 at 21:44
  • @haqnatural I think that should be added to your answer too for completeness. Huge thanks for your answer again. – LearningMath Jun 18 '17 at 21:48
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By definition of (natural) logarithm in $\mathbb{R}$ we have that $\log (x-3)^2=y$ means that $e^y=(x-3)^2$, so, using the definition (in $\mathbb{R}$) of the square root, we have:

$$ e^{\frac{y}{2}}=\sqrt{(x-3)^2}=|x-3|$$

and, going back:

$$ \frac{y}{2}=\log|x-3| \quad \Rightarrow \quad y=\log(x-3)^2=2\log|x-3| $$

Emilio Novati
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The logarithm and non-integer powers are multivalued functions on complex numbers. The general definition for $x \ne 0$ is $x^p = \exp(p \log x)$, where $\log x$ is any branch of the logarithm of $x$, i.e. any $t$ such that $\exp(t) = x$. Thus $p \log x$ is always a logarithm of $x^p$, but in general there may be others not of that form: the others are $2 \pi i n + p \log(x)$ for integers $n$, and this is $p$ times a logarithm of $x$ if and only if $n/p$ is an integer.

Robert Israel
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