How do you solve #log (x + 3) = log (6) - log (2x-1)#?

I'm assuming that all the #log# functions have the same basis #> 1#.

First of all, let's compute the domain of the logarithmic expressions.
For the first one, #x + 3 > 0# must hold, so # x > -3#.
For the last one, we get #2x -1 > 0 <=> x > 1/2#.

As the second condition is the more restrictive one, we can assume that the domain is #x > 1/2# and any possible solutions need to respect this condition.

Now, in order to "get rid" of the logarithmic expressions, it is necessary to simplify the terms on the right-hand side.

To do so, remember the logarithmic rule: #log_a(x) - log_a(y) = log_a(x/y)#

In our case, it means:

#log(x+3) = log(6) - log(2x-1)#
#<=> log(x+3) = log(6/(2x-1))#

Now, due to the fact that for #x#, #y > 0# and #a != 1#, #log x = log y <=> x = y# holds, we can "drop" the logarithms on both sides of the equation.

#<=> x + 3 = 6/(2x-1)#

... multiply both sides with #2x-1#...

# <=> (x+3)(2x-1) = 6#

#<=> 2x^2 +5x - 9 = 0#

This is a quadratic equation that can be solved e. g. with the quadratic formula:

# x = (-b +- sqrt(b^2 - 4ac))/(2a)#

Here, #a = 2#, #b = 5#, #c = -9#, so we can compute the solution as follows:

#x = (-5 +- sqrt(25 - 4 * 2 * (-9))) / 4 = (-5 +- sqrt(97))/4#

The solution # x = (-5 - sqrt(97))/4# is negative, therefore we can ignore it because of our restriction to the domain #x > 1/2#.

So, we have just one solution: # x = (-5 + sqrt(97))/4 ~~ 1.2122#

Hope that this helped!