How do you solve #5+log(2x+1)=6#?

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Answer 1 · 2016-07-16T18:56:21

#x = 4 1/2#

When working with equations which contain logs, each term must be a log term. We cannot work with a mixture of logs and numbers.

#5+log(2x+1)=6#

#log(2x+1)=6 -5#

#log(2x+1)=1#

#log(2x+1)=log 10" if" log A = log B, " then " A = B#

#2x+1 = 10#

#2x = 9 rArr " " x = 4 1/2#