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

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Answer 1 · 2016-02-12T03:54:32

Since the logs are in a common base we can put all logs to one side an simplify by using the logarithm law #log_an - log_am = log_a(n/m)#

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

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

Since nothing is noted in subscript, the log is in base 10.

#10^6 = ((2x + 1)/(x - 1))#

#1 000 000(x - 1) = 2x + 1#

#1 000 000x - 1 000 000 = 2x + 1#

#999 998x = 1 000 001#

#x = (1000001/(999 998))#

Your teacher may want you to keep the answer in fractional form, but he/she may want it rounded off, so keep that in mind.

Practice exercises:

a) #log_7x = 3#

b) #log_3(x + 1) =5 - log_3(4x^2 - 4)#

Good luck!