How do you solve #\log_{7} ( 2x - 3) - \log_{7} ( x + 1) = \log_{7} 2#?

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Answer 1 · 2016-10-23T14:49:46

There is no solution; #{O/}#.

Start by putting all terms to one side of the equation.

#log_7(2x - 3) - log_7(x + 1) - log_7(2) =0#

We will now use the rule #log_a(n) - log_a(m) = log_a(n/m)# to simplify.

#log_7(((2x - 3)/(x + 1))/2) = 0#

#log_7((2x - 3)/(2(x + 1)) )= 0#

Convert to exponential form using the rule #log_a(n) = x -> a^x = n#

#(2x - 3)/(2x+ 2) = 7^0#

#(2x- 3)/(2x+ 2) = 1#

#2x - 3 = 2x + 2#

#0x = 5#

#x = 5/0#

#x = O/#

Hopefully this helps!