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

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Answer 1 · 2015-08-02T07:51:42

Take exponent of both sides to get: #(x+2)(x-1) = 10#

Solve the quadratic and discard the spurious solution to get #x=3#

Note that we require #x+2 > 0# and #x-1 > 0# in order that #log(x+2)# and #log(x-1)# be defined.

This boils down to requiring #x > 1#.

Take exponent of both sides to find:

#10 = 10^1 = 10^(log(x+2)+log(x-1)) = 10^log(x+2)*10^log(x-1)#

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

Subtract #10# from both ends to get:

#0 = x^2+x-12 = (x+4)(x-3)#

Which gives us #x=-4# or #x=3#

Discard the spurious solution #x=-4# since #log(x+2)# and #log(x-1)# are undefined for #x=-4#.

So #x=3#