When the data size and processor speed are both multiplied by 10, then a linearithmic algorithm takes double the time to finish?
I don't like how you expressed this - the ten times faster computer is ten times faster. It will solve the same problem and answer the same question ten times faster. Of course if you give it a question that would take 100 times longer on the old computer, then it will take only 10 times longer on the new computer. 

If you had an algorithm that took n log n microseconds (and in CS "log" usually means log base 2) to solve a problem of size n, a problem of size 10 n would take 10n log (10n) microseconds on the old computer, and n log(10n) microseconds on the new computer. 

log (10n) isn't ten times or twice as much as log n - the factor 10 turns into a constant. log (10n) = log n + log 10 ≈ log n + 3.3. If n = 10 it's twice as much. If n = 10 billion it's only ten percent more. 

Two things to consider: No computer is just "ten times faster". A computer does many different kinds of operations. Some may be ten times faster, some even more, some less. And Big-O is concerned about asymptotic values: I could have an algorithm that takes on my computer n^2 microseconds if n ≤ 1,000,000 and 10n^2 microseconds if n > 1,000,000. So if you want to find actual execution times, you need more than Big-O. 

And it is very common that actual execution times are quite different from what Big-O makes you belief when n is small. For example, let f(n) = n + 1000 log n. f(n) = O(n). But for n ≤ 10,000 f doesn't behave very linear at all.