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: enter image description here

"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 $a^{log_a(x)}=x$?

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    $\begingroup$ That is the definition of $\log_a x$. By definition $b=\log_a x$ is a number $b$ such that $a^b=x$. $\endgroup$ Dec 4 '20 at 14:04
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    $\begingroup$ What the definition of logarithm $\log_a(x)$ do you use? $\endgroup$ Dec 4 '20 at 14:04
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    $\begingroup$ "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. $\endgroup$
    – JMoravitz
    Dec 4 '20 at 14:15
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    $\begingroup$ @JMoravitz yes, it's why I'm posting here now $\endgroup$
    – Doug Fir
    Dec 4 '20 at 14:16
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    $\begingroup$ 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. $\endgroup$ 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,


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!


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