2

I put into Wolfram Alpha:

d/dx 2^x

Where it told me $f'(x)=2^x\log(2)$. Then I put in

d/dx 2^x where x=0

and it said "$\displaystyle \log(2)\approx0.693147$"

I know through Wolfram Alpha and a couple of calculators $\log(2)\approx0.30103$. But Wolfram Alpha and my graphing calculator agree the derivative at $0$ of $2^x$ is $0.693147$ and the log of 2 is $0.30103$. Why?

Michael Hardy
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Gᴇᴏᴍᴇᴛᴇʀ
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  • Two different logarithms differ by their base i.e., e and 10 therefore logarithm of 2 to the base 10 is 0.30103 and to the base e it is 0.693 – Jasser Sep 15 '14 at 03:13

2 Answers2

4

In higher level mathematics, $\log x$ is the natural logarithm, what you know as $\ln x$ or $\log_e x$ and NOT $\log_{10} x$. This is because the number $10$ is almost never necessary compared to $e$... and you probably won't ever see $\log_{10} x$ in the real world, while $\log_e x$ is pretty much everywhere.

Shahar
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    Wolfram Alpha usually states (in grey) in the answer box which logarithm it is using in a particular expression. – user164587 Sep 15 '14 at 03:15
  • @user164587 It does but I doubt many people even notice that. For the most part, however, $\log x$ is the natural logarithm. – Shahar Sep 15 '14 at 03:16
  • My comment was intended to hint to Zack that he should check the WA output again and in full. (I agree about the natural logarithm being the usual one. I always get confused when people use base 10 logarithms without explicitly saying so). – user164587 Sep 15 '14 at 03:20
  • @user164587 Haha, I hadn't ever noticed that. I'm in Calc 1 and we still use log as base 10 if not specified. And most, if not all calculators, naturally use base 10 – Gᴇᴏᴍᴇᴛᴇʀ Sep 15 '14 at 03:44
  • @Zack Are you a freshman ECE major? – Shahar Sep 15 '14 at 03:51
  • @Shahar Yes {need more characters} – Gᴇᴏᴍᴇᴛᴇʀ Sep 15 '14 at 03:54
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Using \begin{align} a^{x} = e^{\ln(a^{x})} = e^{x \ln(a)} \end{align} then \begin{align} \frac{d^{n}}{dx^{n}} \left[ a^{x} \right] = \ln^{n}(a) \, e^{x \ln(a)} = a^{x} \, \ln^{n}(a). \end{align} Evaluating this at $x = 0$ leads to \begin{align} \frac{d^{n}}{dx^{n}} \left[ a^{x} \right]_{x=0} = \ln^{n}(a). \end{align} Taking the case of $n=1$ and $a=2$ provides the value in question.

Leucippus
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