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

1 Answer
Tony B
Nov 18, 2015

# x=-4 " or " +3#

Explanation:

Let "log base" be b

Known that #log_b(b)=1#

Assumption: The use of 'log' in the question is referring to logs to base 10 giving:

#log_10(x+2) +log_10(x-1) =1.....................(1)#

But #color(white)(xxxxxx)log_10(10)=1 .............................(2)#

Substitute (2) into (1) giving:

#log_10(x+2)+log_10(x-1) = log_10(10) ..........(3)#

But #log_10(x+2)+log_10(x-1) = log_10[ (x+2)(x-1)]#

Rewrite (3) as:

#log_10[(x+2)(x-1)] = log_10(10)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Personal note: I am of 'very' old school where #"antilog" -> log^(-1)#

Taking antilogs #-> (x+2)(x-1)=10#

#=> x^2+x-12=0#

#(x +4)(x-3)=0#

#color(blue)( x=-4 " or " +3)#

Related questions
See all questions in Logarithmic Models
Impact of this question
11538 views around the world
You can reuse this answer
Creative Commons License