How do you solve #log_2( 2x) + log _ 2 x = 5#?

1 Answer
Nov 27, 2015

I found #x=4#

Explanation:

We can use the property of logs relating the sum with the product of the arguments and write:
#log_2(2x*x)=5#
use the definition of log:
#2x^2=2^5#
#x^2=32/2#
#x=+-sqrt(16)=+-4#
only the positive one is acceptable.