SOLUTION: Hi. My name is Allen and I've been having difficulty solving these log problems. Here is the problem. Thank you. 1/(2^x)=2^y x = ________

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hi. My name is Allen and I've been having difficulty solving these log problems. Here is the problem. Thank you. 1/(2^x)=2^y x = ________      Log On


   

Question 708956: Hi. My name is Allen and I've been having difficulty solving these log problems. Here is the problem. Thank you.
1/(2^x)=2^y
x = ________
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
A good fact to remember is:
" logs are exponents " That will help you
keep track of them.
+1%2F%282%5Ex%29+=+2%5Ey+
A gerneral rule you can use here is:
+1%2F%28+a%5Eb%29+=+a%5E%28-b%29+
Applying the rule:
+2%5E%28-x%29+=+2%5Ey+
The only way for this to be true is if
+-x+=+y+ or, what is the same,
+x+=+-y+
The theory of logs isn't really needed here
but +-x+ and +y+ are both
logs to the base +2+