Question

What do you call that last function?
y=3x..... linear
\(\large y=3^x \)..... exponential
\(\large y=x^x \) , x>0.... ??????

then again, what's this?
\(\large y=x^{x^x} \) , x>0.... ??????

Answer 1

Also "exponential", base x. Agreed?

Answer 2

If you were going to undo it, log\(_x\) ( )

Answer 3

so this function "grows" exponentially? doesn't it grow faster?

Answer 4

OMG! So I had this nice long explanation and then I hit the backspace and Firefox went "back" and that cleared out everything I wrote. GRRR!

*sigh*

Ok doing this again, you can have term based exponents or constant based exponents but they are still exponential regardless, growth doesn't disqualify it as an exponential.

Proof?

2\(^x\) versus 10\(^x\) versus A3F80C\(^x\) (that's hexidecimal for 10745868)


It doesn't change how the function works, just it's rate of growth. :-)

Recall properties of like terms with exponentials:
(x\(^a\))\(^b\) = x\(^{a•b}\)
x\(^a\)•x\(^b\) = x\(^{a+b}\)
Those back up what I'm describing

And then remember that x\(^x\) when x=0 is 0\(^0\)=1, so

Where this becomes an issue is when x<0 for x\(^x\) because then you have to be careful because (x)\(^x\) alternates back and forth between + and -, agreed?