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Is there a base $b$ such that:

$$\log_b x = x $$

(The only one that comes to mind would be the invalid case of $\log_1 1 = 1 $.)

I'm fairly certain the answer is no, but I can't find a clear justification for it.

(I don't have a strong mathematical background so an answer with the intuition would be much more helpful than any complex theorem proof.)

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    $\begingroup$ The question body is not an appropriate place for any answer. If you want to summarize the other answers, do it as an answer. Make it community wiki if you don't want to feel like stealing credit. $\endgroup$ – Nij Mar 9 at 3:33
  • $\begingroup$ Good point, I didn't know I could answer my own question. I'll fix it, the accepted answer shouldn't lose visibility with this text added a posteriori. I don't feel however as stealing credit from an answer I've accepted and whose author I'm mentioning at the beginning of the text. $\endgroup$ – schomatis Mar 9 at 3:39
  • $\begingroup$ Changing the base of the logarithm just scales it. $\endgroup$ – user76284 Mar 10 at 2:28
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For a function to be a logarithm, it should satisfy the law of logarithms: $\log ab = \log a + \log b$, for $a,b \gt 0$. If it were the identity function, this would become $ab = a + b$, which clearly is not always true.

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Note that $$\log_b x=x\iff b^x=x\iff b=\sqrt[x]{x}.$$ Since $\sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.

But it can be true for some particular $x$. For example $b=\sqrt{2}$ and we have $$\log_{\sqrt{2}}2=2.$$

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No, it can't. For any base $b$, there is some real constant $C$, s.t. $$ \log_b x = C \ln x $$ If it were that this logarithm is identity function, then natural logarithm would be just $x/C$, which is clearly false.

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  • $\begingroup$ If log_b X = C ln x and log_b is the identity function, then log_b x = x = C ln x ==> ln x = x/C . $\endgroup$ – Adam Mar 9 at 16:05
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    $\begingroup$ Further, this shows that the identity function is not even any sort of limit of logarithm functions. $\endgroup$ – R.. Mar 9 at 22:26
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If $b^k = k$ for all $k$ then

$(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.

....

Actually the heck with it: $\log_b 1 = 0$ always and $1 \ne 0$.

Likewise $\log_b b = 1$ and presumably $b \ne 1$

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In general $$log_b a=c$$ is the same as $$b^c=a$$ so you can leave logs behind and focus on solutions to $$b^x=x$$

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I'm expanding here the accepted answer by FredH in the way I interpreted it, putting it in layman's terms as much as possible (according to my very basic mathematical knowledge). I'm also drawing from other answers posted, so thanks everyone who took the time!

First, as suggested, instead of talking in terms of logarithms I'll rephrase the question as an exponentiation (which seems simpler to grasp). In that case we would be looking for a base $b$ that had an exponent that would be the same as its result:

$$ b^x=x $$

In this explanation I'll invert the terms calling the exponent the input and the result of the exponentiation the output (in the logarithm it's actually the other way around since it is the inverse function, but since we're talking about an identity function the relationship between the input and output is symmetrical).

An identity function, and hence a linear function, will have its output growing at the same rate as the input (called both $x$ here). Even if there's a particular $b$ and a particular $x$ that holds the equality (which there are, as shown in the examples of other answers), when $x$ starts growing, say, one unit at a time, in the left-hand side (LHS) it would mean multiplying the already $b^x$ by another $b$

$$ b^{x+1} = x + 1 \iff b^x * b = x + 1 $$

so the LHS would be expanding $b$ times (multiplication) while the output is always growing by a fixed constant of $1$ (addition).

I originally thought of the invalid logarithm base of $1$, but not even $1$ would work for every $x$, it would keep the LHS steady while the output changes. If $b$ is bigger than one, no matter how close to $1$ it is, repeatedly multiplying by it would cause the LHS to, at some point, start (and keep) growing in each iteration by more than just $1$, surpassing the pace of the output (which will always grow at same rate). Worse yet, if $b$ is smaller than one it would mean that the LHS would start getting smaller (while the output grows).

As succinctly expressed in the answers, the identity function, or any linear function for that matter, won't fit a logarithm no matter the chosen base, since it would be like trying to fit a multiplication into an addition.

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Logically, $y=x$ is a straight line, $y=\log_b x$ is not (otherwise why would we call it non-linear function) so they cannot coincide for all $x$.

Suppose $y=\log_b x$ is a straight line the same as $y=x$ then for any two $x_1$ and $x_2$ such as $x_2=x_1+1$ we should have for the slope: $1=\frac{\log_b x_2-\log_b x_1}{x_2-x_1}=\log_b x_2-\log_b x_1=\log_b \frac{x_2}{x_1} \rightarrow \frac{x_2}{x_1}=b$ which leads to a contradiction ($\frac{4}{3}=b, \frac{5}{4}=b$).

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    $\begingroup$ And why is $y=\log_b x$ not a straight line? $\endgroup$ – Henning Makholm Mar 8 at 22:26
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    $\begingroup$ @HenningMakholm: because it does not have a constant slope (derivative will be $c\over x$ where $c$ is a constant but $x$ is obviously not) $\endgroup$ – Vasya Mar 8 at 22:32
  • $\begingroup$ $\frac0x$ is a constant function on a useful subset of $\mathbb R$. $\endgroup$ – Henning Makholm Mar 8 at 22:37
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    $\begingroup$ @HenningMakholm: we both know that $c \ne 0$ in this case, I was trying to explain it at the simplest level for the original poster. $\endgroup$ – Vasya Mar 9 at 3:23
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    $\begingroup$ The OP was asking for an argument, not for a "simplest level" that amounts to an unsupported assertion of the same thing he wanted a proof of. $\endgroup$ – Henning Makholm Mar 9 at 10:14

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