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!

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