How do you solve #log_2x+log_2 3=3#?

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Answer 1 · 2016-08-21T03:53:24

I found: #x=8/3#

You can condense the two logs and write:
#log_2(3x)=3#
use the definition of log and write:
#3x=2^3#
#3x=8#
#x=8/3#

Answer 2 · 2016-08-21T04:03:33

#x=8/3#

As #2^3=8#, we have #log_2 8=3#, hence

#log_2 x+log_2 3=3=log_2 8# or

#log_2(3×x)=log_2 8# or

#3x=8# or

#x=8/3#