Sorry, this is a really simple question, but I'm trying to teach myself calculus and can't figure it out. If we define $\log(b) = \frac{db^x}{dx}(0)$ how does one prove $\log(b^a) = a\log(b)$? I tried using the definition of derivative but got stuck at $\lim_{h \to 0}\frac{b^{ah} - 1}{h}$.
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This will help: http://math.stackexchange.com/questions/331153/proving-that-lim-h-to-0-fracbh-1h-lnb – Vishwa Iyer Jul 21 '14 at 22:14
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1There is no calculus in the logarithm definition I know. Please read this http://en.wikipedia.org/wiki/Logarithm. Just the Definition part. – nitzpo Jul 21 '14 at 22:16
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1Yes, but I just said I'm using a differentiated definition. – Elliot Gorokhovsky Jul 21 '14 at 22:19
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That's the whole point. Otherwise it's trivial. – Elliot Gorokhovsky Jul 21 '14 at 22:19
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@RenéG: You can also put a hammer to this problem with the chain rule: $d(b^a)^x/dx = db^{ax}/dx = (dax/dx)(db^{ax}/dax)$ which taken at $x=ax=0$ gives $a\log b$. – Eric Stucky Jul 21 '14 at 22:21
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Ya, I feel stupid now I should have tried that – Elliot Gorokhovsky Jul 21 '14 at 22:22
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If $a\not=0$, let $k=ah$ and note that $h\to0$ iff $k\to0$, which gives
$$\lim_{h\to0}{(b^a)^h-1\over h}=a\lim_{h\to0}{b^{ah}-1\over ah}=a\lim_{k\to0}{b^k-1\over k}$$
If $a=0$, then $(b^a)^h-1=0$, so the limit is obviously $0$.
Barry Cipra
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Hint: $\displaystyle \log(b^a) = \frac{d(b^{ax})}{dx}(0)$ and using the chain rule $$\frac{d}{dx}b^{ax} = \frac{d}{dx}(b^{x})^a = a(b^x)^{a-1}\frac{d}{dx}b^{x}$$
Mathmo123
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Using $(b^a)^x = b^{ax}$ and the chain rule, we get \begin{align} \log(b^a) &= \left.\frac{\mathrm d}{\mathrm dx} b^{ax}\right|_{x=0} = \left.\left(\frac{\mathrm d}{\mathrm dy} b^y\right)\right|_{y=0} \cdot\left.\left( \frac{\mathrm d}{\mathrm dx} ax\right)\right|_{x=0} = \log(b)\cdot a. \end{align}
Christoph
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If $x = 0$ then wouldn't the derivative of $ax$ be 0? I mean, the function $0a$ is flat. – Elliot Gorokhovsky Jul 21 '14 at 22:33
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You might want to change that to be consistent with the other answers. – Elliot Gorokhovsky Jul 21 '14 at 22:35
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The notation $\left.\left(\frac{\mathrm d}{\mathrm dx} \dots\right)\right|_{x=x_0}$ is interpreted as "first take the derivative of $\dots$ with respect to $x$, then evaluate at $x=x_0$. It is very common and in my opinion less confusing than $\frac{\mathrm d(\dots)}{\mathrm dx}(x_0)$. – Christoph Jul 21 '14 at 22:50
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$$ \log(a^b) = \int_1^{a^b} \frac{dx}x. $$ Suppose $w^{\,b}=x$. Then $bw^{b-1}\,dw= dx$, so $b\dfrac{dw}{w} = \dfrac{dx}x$. As $x$ goes from $1$ to $a^b$, then $w$ goes from $1$ to $a$. Hence, $$ \log(a^b) = \int_1^{a^b} \frac{dx}x = \int_1^a b \frac{dw}w = b\log a. $$
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I did the same thing. The comment from the OP was "That's not the definition I am using." There is no accounting for taste. I then deleted mine due to downvotes, of course. – Will Jagy Jul 21 '14 at 22:34
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I haven't closely examined whether there might be some circularity. I should be OK if $b$ is an integer or even if it's rational, but the proof that $(d/dw)w^b=bw^{b-1}$ when $b$ is irrational may involve logarithms and the chain rule. Although I think maybe it could be done by a squeezing argument. – Michael Hardy Jul 21 '14 at 22:37
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For that matter, isn't the definition of $b^x$ $e^{x\log{b}}$ which kind of depends on the properties of log in the first place.... that being said, I'm agreed that this answer doesn't really answer the question – Mathmo123 Jul 21 '14 at 22:42