I saw very often that:
$$\log_{10}(x) = \frac{\ln(x)}{\ln(10)}$$
So I assume it's true for every base:
$$\log_{a}(x) = \frac{\ln(x)}{\ln(a)}$$
But, is the following statement true?
$$\log_{a}(x) = \frac{\log_n(x)}{\log_n(10)}$$
Where $n$ is an arbitrary non-zero real number.
On
The statement is not true.
Choose $a = 1$. Then $\log_b(a) = 0$ for all $b > 0$ and $b \neq 1$. Therefore, $\log_a(x) = \dfrac{\log_b(x)}{\log_b(a)}$ would not be defined.
On
The answer is: "no".
The second expression is correct for any $a$ with $a>0$ and $a\neq1$.
Observe: if the third expression would be true then combination with the second gives: $$\ln(a)=\log_n(10)$$ which is of course not true for every legal base $a$. The RHS is fixed, but the LHS is not fixed.
edit (to verify your assumption that the second expression is true)
The expression $\log_{a}x$ only makes sense if $x>0$, $a>0$ and $a\neq1$.
It can be interpreted as the unique number that satisfies:
$$a^{\log_{a}x}=x$$ Special case $a=e$ gives: $$e^{\ln x}=x$$ Now observe that:
$$a^{\frac{\ln x}{\ln a}}=\left(e^{\ln a}\right)^{\frac{\ln x}{\ln a}}=e^{\ln a\times\frac{\ln x}{\ln a}}=e^{\ln x}=x$$
This allows us to conclude that: $$\log_{a}x=\frac{\ln x}{\ln a}$$
Assuming $a, n > 0$ and $a\neq 1$ indeed $$\log_a n = \frac{\ln n}{\ln a},$$
this is because $a^x = (e^{\ln a})^x = e^{x\cdot\ln a}$, so $x\cdot\ln a = \ln n$. Yet, $e$ here isn't any special number except that $\ln e = 1$. Thus for any $b > 0$, $b\neq 1$ $$a^y = (b^{\log_b a})^y = e^{y\cdot\log_b a},$$ so $y\cdot\log_b a = \log_b n$. Also, using the first formula we can derive
$$\frac{\log_b n}{\log_b a} = \frac{\frac{\ln n}{\ln b}}{\frac{\ln a}{\ln b}} = \frac{\ln n \cdot \ln b}{\ln a \cdot\ln b} = \frac{\ln n}{\ln a} = \log_a n.$$
Intuition:
You can think intuitively of an logarithm as of length of a number (i.e., the number of digits in some base $b > 1$ representation). If your number is $$\underbrace{100\ldots00_b}_{k\text{ digits}}=b^k,$$ then precisely $k = \log_b n$.
Now, if you want to change base $b$ to $b^2$, then it's enough to "pair up" your digits,
because a base-$b$ $2$-digit number is equivalent to a base-$b^2$ $1$-digit number. Therefore, the length of your number in new units is 2 times shorter, only now each "digit" takes 2 units of space instead of just one.
$$\mathtt{156}_{10} = {\color{red}{\mathtt{1}}}{\color{darkred}{\mathtt{0}}} {\color{blue}{\mathtt{0}}}{\color{darkblue}{\mathtt{1}}} {\color{red}{\mathtt{1}}}{\color{darkred}{\mathtt{1}}} {\color{blue}{\mathtt{0}}}{\color{darkblue}{\mathtt{0}}}_2 = ({\color{red}{\mathtt{1}}}{\color{darkred}{\mathtt{0}}}) ({\color{blue}{\mathtt{0}}}{\color{darkblue}{\mathtt{1}}}) ({\color{red}{\mathtt{1}}}{\color{darkred}{\mathtt{0}}}) ({\color{blue}{\mathtt{0}}}{\color{darkblue}{\mathtt{0}}} )_{({\color{red}{2}}\times{\color{blue}{2}})} = {\color{red}{\mathtt{2}}} {\color{blue}{\mathtt{1}}} {\color{red}{\mathtt{3}}} {\color{blue}{\mathtt{0}}}_4$$
In other words $\log_a n = \frac{\log_b n}{\log_b a}$ which is equivalent to $\log_b a = \frac{\log_b n}{\log_a n}$ means that ratio of base-$b$ length to base-$a$ length is equal to how long is $a$ in $b$. That is, if $a$ takes $3$ digits in base $b$, then $n$ takes $3$ times more base-$b$ digits than base-$a$ digits.
I hope this helps $\ddot\smile$