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

1 Answer
George C.
Oct 14, 2016

Using the Complex logarithm:

#x = -2# or #x = -3#

Explanation:

Add #ln(x+1)# to both sides to get:

#ln(x-1) = ln(x+5)+ln(x+1)#

#color(white)(ln(x-1)) = ln((x+5)(x+1))#

#color(white)(ln(x-1)) = ln(x^2+6x+5)#

Take the exponent base #e# of both ends to get:

#x-1 = x^2+6x+5#

Subtract #x-1# from both sides to get:

#0 = x^2+5x+6 = (x+2)(x+3)#

So #x = -2# or #x = -3#

Note that these values result in taking logarithms of negative values.

So the original equation has no solutions using the Real valued logarithm.

However, it does give us solutions using the Complex logarithm:

#ln((color(blue)(-2))-1) - ln((color(blue)(-2))+1) = ln(-3) - ln(-1)#

#color(white)(ln((color(white)(-2))-1) - ln((color(white)(-2))+1)) = (ln(3) + pii) - (ln(1) + pii)#

#color(white)(ln((color(white)(-2))-1) - ln((color(white)(-2))+1)) = ln(3)#

#color(white)(ln((color(white)(-2))-1) - ln((color(white)(-2))+1)) = ln((color(blue)(-2))+5)#

#ln((color(blue)(-3))-1) - ln((color(blue)(-3))+1) = ln(-4) - ln(-2)#

#color(white)(ln((color(white)(-3))-1) - ln((color(white)(-3))+1)) = (ln(4) + pii) - (ln(2) + pii)#

#color(white)(ln((color(white)(-3))-1) - ln((color(white)(-3))+1)) = ln(4/2)#

#color(white)(ln((color(white)(-3))-1) - ln((color(white)(-3))+1)) = ln(2)#

#color(white)(ln((color(white)(-3))-1) - ln((color(white)(-3))+1)) = ln((color(blue)(-3))+5)#

So both #x=-2# and #x=-3# are solutions.

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