### Instruction: What is the domain of the function $f(x)=\log_2(\log_3(\log_4(\log_5x)))$?### Response: To find the domain of a function, I need to consider what values of $x$ make the function well-defined and real-valued.
A logarithm function $\log_ba$ is only defined when $a>0$ and $b>0$, and $b\neq 1$.
So, for the innermost logarithm, $\log_5x$, I need $x>0$ and $5>0$, and $5\neq 1$.
The last two conditions are always true, so I only need to worry about $x>0$.
For the next logarithm, $\log_4(\log_5x)$, I need $\log_5x>0$ and $4>0$, and $4\neq 1$.
The last two conditions are always true, so I only need to worry about $\log_5x>0$.
But when is $\log_5x>0$?
That means $5^{\log_5x}>5^0$, or $x>1$.
So, I need $x>1$ for the second logarithm to be defined.
For the third logarithm, $\log_3(\log_4(\log_5x))$, I need $\log_4(\log_5x)>0$ and $3>0$, and $3\neq 1$.
The last two conditions are always true, so I only need to worry about $\log_4(\log_5x)>0$.
But when is $\log_4(\log_5x)>0$?
That means $4^{\log_4(\log_5x)}>4^0$, or $\log_5x>1$.
So, I need $\log_5x>1$ for the third logarithm to be defined.
But when is $\log_5x>1$?
That means $5^{\log_5x}>5^1$, or $x>5$.
So, I need $x>5$ for the third logarithm to be defined.
For the fourth and final logarithm, $\log_2(\log_3(\log_4(\log_5x)))$, I need $\log_3(\log_4(\log_5x))>0$ and $2>0$, and $2\neq 1$.
The last two conditions are always true, so I only need to worry about $\log_3(\log_4(\log_5x))>0$.
But when is $\log_3(\log_4(\log_5x))>0$?
That means $3^{\log_3(\log_4(\log_5x))}>3^0$, or $\log_4(\log_5x)>1$.
So, I need $\log_4(\log_5x)>1$ for the fourth logarithm to be defined.
But when is $\log_4(\log_5x)>1$?
That means $4^{\log_4(\log_5x)}>4^1$, or $\log_5x>4$.
So, I need $\log_5x>4$ for the fourth logarithm to be defined.
But when is $\log_5x>4$?
That means $5^{\log_5x}>5^4$, or $x>625$.
So, I need $x>625$ for the fourth logarithm to be defined.
So, putting it all together, I need $x>625$ for the function $f(x)$ to be defined and real-valued.
This is the domain of the function.