How do you solve # ln(x-4)-ln(3x-10)=ln (1/x)#?

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Answer 1 · 2016-09-07T13:44:57

#x= 5#

Use the rule #log_a(n) - log_a(m) = log_a(n/m)#.

#=>ln((x - 4)/(3x - 10)) = ln(1/ x)#

Now use the rule #lna = lnb-> a = b#.

#(x - 4)/(3x - 10) = 1/x#

#x^2 - 4x = 3x - 10#

#x^2 - 7x + 10 = 0#

#(x - 5)(x - 2) = 0#

#x = 5 and 2#

However, #x = 2# is extraneous, since it renders the equation undefined.

Hopefully this helps!