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

1 Answer
Noah G
Dec 1, 2016

#O/#

Explanation:

Use the subtraction rule of logarithms that #log_a(n) - log_a(m) = log_a(n/m)#.

#ln(x/(x + 1)) = 1#

#x/(x + 1) = e^1#

#x = e(x + 1)#

#x = ex + e#

#x- ex = e#

#x(1 - e) = e#

#x= e/(1 - e)#

If you want an approximation, #x ~= -1.58#. However, since #x > 0# for the natural logarithms to be defined, this solution is extraneous.

Hopefully this helps!

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