How do you solve #log(x-1)+log(x+1)=2 log(x+2)#?

1 Answer
mason m
Jan 19, 2016

There are no solutions.

Explanation:

Use the logarithm rules to simplify either side:

  • Left hand side: #loga+logb=log(ab)#
  • Right hand side: #bloga=log(a^b)#

This gives

#log[(x-1)(x+1)]=log[(x+2)^2]#

This can be simplified using the following rule:

  • If #loga=logb#, then #a=b#

Giving us:

#(x-1)(x+1)=(x+2)^2#

Distribute both of these.

#x^2-1=x^2+4x+4#

Solve. The #x^2# terms will cancel, so there will only be one solution.

#4x=-5#

#x=-5/4#

However, this solution is invalid. Imagine if #x# actually were #-5/4#. Plug it into the original equation. The terms #log(x-1)# and #log(x+1)# would be #log(-9/4)# and #log(-1/4)#, and the logarithm function #loga# is only defined when #a>0#.

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