# Why is $a^x = e^{x \log a}$?

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)?

• In real analysis, $a>0$. – Wuestenfux Dec 24 '18 at 14:49
• That is a definition of $a^x$. – Bernard Dec 24 '18 at 15:36

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

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

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

$$\ln(x)$$ 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 $$a^x=e^{\ln(a^x)}=e^{x\ln(a)}$$

• Use a backslash in front of $\ln$ to render it properly. – Don Thousand Dec 24 '18 at 14:56
• thank you, didn't know about it – user531476 Dec 24 '18 at 14:58

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

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

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, $$e^{\ln a^x} = e^{x\ln a}$$.

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.