# Why does $a^{\log_a(x)}=x$?

I'm currently learning about logs and yesterday made a post on here.

In that post I was told that $$a^{\log_a(x)}=x$$

From a commenter: "It should be clear that..."

Nope, not for me. I've tried coming back to this since posting yesterday but cannot 'see it' or make it click. I'm seeking hand holding and a low level response. Why does $$a^{log_a(x)}=x$$?

• That is the definition of $\log_a x$. By definition $b=\log_a x$ is a number $b$ such that $a^b=x$. Dec 4 '20 at 14:04
• What the definition of logarithm $\log_a(x)$ do you use? Dec 4 '20 at 14:04
• "using a calculator I can see that..." There's your problem right there. You've relied too heavily on calculators and never learned what the symbols you are pressing actually mean. Dec 4 '20 at 14:15
• @JMoravitz yes, it's why I'm posting here now Dec 4 '20 at 14:16
• When I was young, I was taught that, in algebra, in addition to the already known four arithmetic operations, we are introducing three more: exponentiation, radicals/roots and logarithms. If $x^y=z$, knowing $x$ and $y$, you find $z$ by exponentiation. Knowing $y$ and $z$, you find $x$ using roots ($x=\sqrt[y]{z}$). Knowing $x$ and $z$, you find $y$ using logarithms: $y=\log_x(z)$. Later I learned a lot more (in particular that roots are not much different from exponentiation) - but I still think this is quite a good way to introduce logarithms and where they pop up in maths. Dec 4 '20 at 14:28

By definition, the $$\log-$$function is the inverse of the exponential function. It means that, if $$f:\mathbb R\to \mathbb R^+$$ such that, $$f(x)=a^x$$ then its inverse is a function $$f^{-1}:\mathbb R^+\to \mathbb R$$ such that,

$$f(f^{-1}(x))=x.$$

We then define $$f^{-1}$$ as $$f^{-1}(x)=\log_a x$$. So,

$$f(f^{-1}(x))=x\Leftrightarrow a^{\log_a x}=x.$$

It is true that by definition $$a^{\log_a(x)}=x$$. However, perhaps it would be worthwhile to show you the motivation behind why we define logarithms in this way. As @Stinking Bishop has already pointed out, logarithms help us solve for $$x$$ when dealing with an equation of the form $$a^x=b \, .$$ For example, if $$10^x=100$$ then $$x=2$$ is the solution. But suppose we were trying to solve $$10^x=101 \, .$$ Now, it is not immediately clear what $$x$$ is. We can approximate its decimal value using trial and error, but this is only an approximation. In the same way that $$\sqrt{2}$$ represents the exact positive solution of $$x^2=2$$, $$\log_{10}(101)$$ represents the exact solution to the equation $$10^x=101 \, .$$ This means that $$10^{\log_{10}(101)}=101$$ by definition. $$\log_{10}(101)$$ is the 'label' we give to the solution of $$10^x=101 \, .$$ More generally, if $$a^x=b$$, then $$\log_a{b}$$, by virtue of how logarithms are defined, represents the solution of the equation. Thus, $$a^{\log_a{b}}=b$$ by definition. Now, a more sophisticated way of thinking about logarithms is that they inverse exponentiation. If $$f(x)=a^x$$, then $$f^{-1}(x)=\log_a(x)$$. The defining feature of inverse functions is that $$f^{-1}(f(x))=x$$ and $$f(f^{-1}(x))=x$$ for all $$x$$. Hence, $$\log_a(a^x) = x \text{ and } a^{\log_a{x}}=x$$ are both true by definition. The hardest part is trying to explain why these two conceptions of logarithms, while superficially different, are actually the same. Let's return to our equation $$a^x=b \, .$$ If we 'take logs of both sides', we get $$\log_a(a^x) = \log_a(b) \, ,$$ which simplifies to $$x= \log_a(b).$$ Thus, by defining the logarithm as the inverse of the exponential, $$x$$ represents the power that we must raise $$a$$ to in order to get $$b$$. And this should make intuitive sense. In general, the inverse function should tell us 'how to go backwards'. If we start with the number $$x$$ and perform a function on it so that we get $$f(x)$$, then $$f^{-1}$$ is like an instruction manual for how to get back to $$x$$. Thus, $$f^{-1}(f(x))=x \, .$$ In the case of exponentiation, we start with a number with a number $$x$$, and raise $$a$$ to the power of $$x$$ to get $$a^x$$. The logarithm, being the inverse function, should let us return to $$x$$: $$\log_a(a^x)=x \, .$$ Also, if we first perform $$f^{-1}$$ on $$x$$, then again, $$f$$ is the inverse function. Try working out what this means in the context of exponentials and logarithms.

Consider the statement $$\color{red}{2}^\color{green}{3}=\color{blue}{8}$$

You can say that:

• $$\color{green}{3}=\log_\color{red}{2}(\color{blue}{8})$$
• $$\color{red}{2}=\sqrt[\color{green}{3}]{\color{blue}{8}}$$

Instead of asking "what number must $$\color{red}{2}$$ be raised to, to give $$\color{blue}{8}$$" you may ask "what is $$\log_\color{red}{2}(\color{blue}{8})$$?". You can then consider $$\color{red}{a}^{\log_\color{red}{a}(\color{blue}{x})}$$

$$\log_\color{red}{a}(\color{blue}{x})$$ is "the number that gives $$\color{blue}{x}$$ when $$\color{red}{a}$$ is raised to it", so Очевидно, что $$\color{red}{a}^{\log_\color{red}{a}(\color{blue}{x})}=\color{blue}{x}$$

In general (given that both $$\log_a(c), \sqrt[b]c$$ exist) if $$\color{red}{a}^\color{green}{b}=\color{blue}{c}$$ then

• $$\color{green}{b}=\log_\color{red}{a}(\color{blue}{c})$$, which means that $$\color{red}{a}^{\log_\color{red}{a}(\color{blue}{c})}=\color{blue}{c}$$

• $$\color{red}{a}=\sqrt[\color{green}{b}]{\color{blue}{c}}$$, which means that $$(\sqrt[\color{green}{b}]{\color{blue}{c}})^{\color{green}{b}}=\color{blue}{c}$$

I recommend watching this.

Assume that $$a^b = x$$. If we take the logarithm of both sides in base $$a$$, we will get $$\log_a(a^b) = \log_a(x)$$. By the definition of logarithm, $$log_a(a^b) = b$$, since logarithm function returns the exponent of an input number in the given base. Thus, returning to $$\log_a(a^b) = \log_a(x)$$, we can see that $$b = \log_a(a^b) = \log_a(x)$$, and $$b = \log_a(x)$$. Since our original equation is $$a^b = x$$, we can just substitute $$b$$ in the original equation to find the expression $$a^{\log_a(x)} = x$$. Hope this is clear!