Is there a logarithm base for which the logarithm becomes an identity function?

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.)

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.

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 $$\log_{\sqrt{2}}2=2.$$

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.

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 $b \ne 1$