find the inverse of 2^x

Question

anonymous · Answer

I figure i have to switch my x and y  so $$x=2^{y}$$

anonymous · Answer

should I ln or log both sides or what?

pooja195 · Answer

Log(y) / Log(2)

zepdrix · Answer

To get rid of the exponential base 2, you'll want to take the log base 2 of both sides :D it's a little awkward. But after you do that, you can do the log trick that pooja mentioned.

anonymous · Answer

so $$\log_{2}x=\log_{2}2^{y}  $$

pooja195 · Answer

yep

anonymous · Answer

I dont get pooja's trick

anonymous · Answer

what should i do from there?

anonymous · Answer

well i know the right side would kinda cancel so i have $$\log_{2}x=y $$

zepdrix · Answer

Changing the base of a log, here is example of the rule,$$\large \log_a (b)=\frac{\ln b}{\ln a}$$You can make the new base whatever you want, I did e in this case (the most natural of all logs :D)

anonymous · Answer

Im not seeing why you have to do that or what i'm really suposed to be doing. i"m kinda confused. sorry

zepdrix · Answer

$$\large \log_2 x =\frac{\log_c x}{\log_c 2}$$It's a rule that lets us change our base to any new value $c$ that we want.
It's just a rule to remember :O I can't remember how it's derived actually...

anonymous · Answer

why do i need to change my base?

anonymous · Answer

i have $$\log_{2}x=y $$

zepdrix · Answer

I suppose you don't! :)
But the log base 2 is fairly insignificant. It's not a value that can be computed easily.

Yah that's a fine answer ^^ prolly stop there. my bad

anonymous · Answer

oh so you are changing the base so the answer seems less trivial?

zepdrix · Answer

Yah, which probably isn't necessary for your class XD hehe

anonymous · Answer

alright. awesome!