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Why is $ a^x = e^{x \log a}$, where $ a $ is a constant?

From my research, I understand that the the natural log of a number is the constant you get when you differentiate the function of that number raised to the power $ x$. For example, if we differentiate the function $ 2^x $, we get $ 2 \log 2$. I also understand that the natural log of e is just 1. But I cannot connect the dots here.

I would really appreciate an intuitive explanation of why we can write a number raised to a power as e raised to (the power x the natural logarithm of the number)?

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  • $\begingroup$ In real analysis, $a>0$. $\endgroup$ – Wuestenfux Dec 24 '18 at 14:49
  • $\begingroup$ That is a definition of $a^x$. $\endgroup$ – Bernard Dec 24 '18 at 15:36
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I think we can agree that

$$a=e^{\log a}$$

which arises from one of the properties of the logarithm. Therefore, it’s sufficient to say that

$$a^x=e^{\log a^x}$$

But one of the properties of the logarithm also dictates that

$$\log a^x=x\log a$$

Therefore

$$a^x=e^{x\log a}$$

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$$e^{x\log a}=e^{\log a^x}=a^x$$

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$\ln(x)$ is the inverse function of $e^x$ and so we have $e^{\ln(x)}=x$ moreover, from the properties of $\ln(x)$ is that $\ln(a^b)=b\ln(a)$ so $a^x=e^{\ln(a^x)}=e^{x\ln(a)}$

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    $\begingroup$ Use a backslash in front of $\ln$ to render it properly. $\endgroup$ – Don Thousand Dec 24 '18 at 14:56
  • $\begingroup$ thank you, didn't know about it $\endgroup$ – user531476 Dec 24 '18 at 14:58
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As noted by the other answers, this is due to $e^x$ and $\log x$ being inverses.

However, you can also note that this is a combination of two important properties of logarithms of all bases:

  • $$\log_a b^c = c\log_b$$

  • $$a^{\log_a b} = b$$

The second property is easy to understand. $\log_a b$ means the number $a$ needs to be raised to to get $b$. So raising $a$ to “the number $a$ needs to be raised to give $b$” obviously gives $b$. Mathematically, you could say

$$\log_a b = c \iff a^c = b$$

$$a^{\log_a b} = a^c = b$$

The first property is can be thought of as a repeated addition property:

$$\log_a bc = x \iff a^x = bc$$

$x = \log_a b+\log_a c$ gives

$$\underbrace{a^{\log_a b+\log_a c}}_{a^{\log_ ab+\log_a c} = a^{\log_a b}\cdot a^{\log_a c} = bc} = bc \iff \color{blue}{\log_a(bc) = \log_a b+\log_a c}$$

from which you reach the first property.

$$\log_a (bc) = \log b+\log c \implies \log_a b^c = \underbrace{\log_a b+\log_ab+…+\log_a b}_{c \text{ times}} = c\log_a b$$

Combining the two properties, you get

$$e^{a\log x} = e^{\log a^x} = a^x$$

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The decisive point is that the functions $\exp:{\Bbb R}\rightarrow {\Bbb R}_{>0}:x\mapsto e^x$ and $\log:{\Bbb R}_{>0}\rightarrow {\Bbb R}:x\mapsto \log_ex = \ln x$ are inverse to each other. Then we have $a^x = e^{\ln a^x}$. By the property of the logarithm, $e^{\ln a^x} = e^{x\ln a}$.

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Use that $$\ln(e^p)=p\implies\ln(e^{\ln q})=\ln q\implies e^{\ln q}=q$$ The first result will take you the rest of the way.

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