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We have:

$$\log_a{x^n}=\log_a{\left( x\cdot x \cdot x \cdot ...\cdot x\right)}=\log_a(x)+\log_a(x)+...+\log_a(x)=n\log_a(x)$$

But, $$ \ln\left(-1\right)^2=\ln{\left(-1\right)^2}\\\ln(1)=2\ln(-1)\\ 0=2i\pi$$

or

$$ \log_a(-1)^3=\log_a(-1)^3 \\ \log_a(-1)=3\log_a(-1) \\1=3 $$

So, if the "identity" first written is just for $x\in\mathbb{R}| x>0$ tell me why, or where I'm I failing at!

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5  
The logarithm becomes multivalued when you leave $\mathbb R^+$. So, yes to the first part of your last sentence. –  Rahul Jan 4 '13 at 4:02
1  
See here for more information too. –  JavaMan Jan 4 '13 at 4:04
    

3 Answers 3

up vote 3 down vote accepted

If you'd like to use the identity $$ \ln x^n=n \ln x, $$ you should consider the definition of logarithm as a multivalued function: $$ \textrm{Ln}\, z=\ln |z|+i \mathrm{Arg}\, z=\ln |z|+i(\theta+2\pi k), $$ i.e., there are infinitely many values. Using this definition your examples are valid: $$ \textrm{Ln}\,(-1)^2=\textrm{Ln}\, (-1)^2\\ \textrm{Ln}\,1=2 \textrm{Ln}\, (-1)\\ i 2\pi k=2i(\pi+2\pi m), $$ where the last equality is understood modulo $2\pi k i$.

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The identities

$$\log_a{x^n}=n\log_a(x)$$ and $$\log_a(xy)=\log_a(x)+\log_a(y) \,.$$

are proven for positive $x,y$. Their proof uses (hidden) the fact that the exponential function $a^x : \mathbb R \to (0, \infty)$ is a bijection....This is not true anymore when you extend it to complex numbers.

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The exponentiation function is not one-to-one. For any $a$ and $x$, there are many complex numbers $y$ such that $$a^y = x$$ and therefore there are many numbers $y$ that of which it can be said that $$y=log_a x.$$ For positive real $x$ we can define $y$ to be the unique real number for which this holds; for other $x$, we cannot.

This means that you cannot conclude from $$\log_a x = \log_a x'$$ that $$x = x'$$ except under certain limited circumstances, just as you cannot conclude from $$x^2 = y^2$$ to $$x = y$$ except under certain limited circumstances.

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