How do you solve #Ln(x + 5) + ln(x - 1) = 2#?

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Answer 1 · 2016-06-21T12:56:47

#x = sqrt(9+e^2) -2#.

e = 2.7183, nearly.

#x>1# to make #log (x-1)# real.

Use #ln e^n=n and ln a + ln b = ln ab#..

The given equation is

#ln((x+5)(x-1))= ln e^2#.

So, #(x+5)(x-1)=e^2#. Solving,

#x = -2+-sqrt(9+e^2)#

The negative root is inadmissible.

So, #x = sqrt(9+e^2)-2#.

e = 2.7183, nearly...