How do you solve #log_2 (x+2) - log_2(x) =8#?

1 Answer
Sep 18, 2015

I found: #x=2/255=0.007843#

Explanation:

You can use first a rule of the log that says:
#logx-logy=log(x/y)#
and get:
#log_2[(x+2)/x]=8#
Then you use the definition of log to get:
#(x+2)/x=2^8#
rearranging:
#x+2=256x#
#255x=2#
#x=2/255=0.007843#