While I understand the proof of the power rule for logarithms for positive integers, I can not prove them for fractional and negative powers. Also curious to know what would a proof for irrational powers look like?
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Try it a bit harder. Use $1/x=x^{-1}$. For irrational powers, the key word is continuity. – Berci Dec 29 '13 at 02:31
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If you don't mind using calculus, start with $$ \ln(x) = \int_1^x \frac{dt} t .$$ Make the substitution $t = s^a$, so $dt = a s^{a-1} ds$, and you get $$ \ln(x^a) = \int_1^{x^a} \frac{dt} t = a \int_1^{x} \frac{ds} s = a \ln(x).$$
Stephen Montgomery-Smith
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This is perfect if the logarithm is defined this way, which is what a real analysis course in most universities might have done, and hence this kind of solution is necessary. Personally it does seem to me a completely unnatural definition, since $log_2(x)$ for example has a very simple "elementary" meaning which is obscured if it is defined as $ln(x)/ln(2)$. Note also that we can get all the basic properties of $\log$ without integration if we use the alternative definition, because we can get all the properties of exponentiation by just limits. – user21820 Dec 29 '13 at 05:32
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Since you didn't specify, I assume that logarithms are defined as the inverse of exponentiation. Specifically, $a^{\log_a(x)} = x$ for any $a>1$ and $x>0$, which also gives $\log_a(a^x)=x$ for any $a>1$ and $x \in \mathbb{R}$. It is easy to prove the rule $a^{bc} = (a^b)^c$ for any $a>0$ and $b,c \in \mathbb{R}$, by first dealing with integer $b$,$c$ and then rationals and then approximating irrational by a limiting sequence of rationals. Now apply $\log_a$ to both sides to get $bc = \log_a((a^b)^c)$. Note that $b = \log_a(a^b)$, so if we substitute $b = \log_a(x)$ for $x>0$, we get $c \cdot \log_a(x) = \log_a((a^{\log_a(x)})^c) = \log_a(x^c)$ for any $a>0$ and $x>0$ and $c \in \mathbb{R}$.
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Excellent answer. I am still learning and this was easy to understand and follow. Thanks a lot. – astraldust Dec 29 '13 at 05:53
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You're welcome! Note that for $a \in (0,1)$ we can use $a^x = 1/(1/a)^x = (1/a)^{-x}$ to define $log_a$ to have the same property $a^{log_a(x)} = x$ and thus get the same result. For negative $a$ you will need complex numbers and either multi-valued functions or branch cuts to define the logarithm. Multi-valued functions will still preserve these properties, but are multi-valued. Branch cuts give you proper functions, but these properties won't hold anymore. – user21820 Dec 29 '13 at 06:14
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@CiaPan: Thanks for editing! I don't know why I didn't use the proper LaTeX at that time. – user21820 Feb 01 '16 at 15:32
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You are actually asking if the following true for $a,b\in\mathbb R$. $$(x^a)^b=x^{ab}.$$
For irrational number, you can find an explanation here.
mathlove
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Hmm I think you got it wrong; see my answer. Your identity corresponds to the "product rule" for logarithms, not the one astraldust is asking about. – user21820 Dec 29 '13 at 02:41
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