How do you solve #log(x - 3) = 1 - log(x)#?

1 Answer
Özgür Özer
Dec 3, 2015

#=>x_1=-2, x_2=5#

Explanation:

#log(x+3)=1-log(x)#
#=>log(x+3)+log(x)=1-log(x)+log(x)#
The logarithm of a product is the sum of the logarithms of the numbers being multiplied:
#log(x^2-3x)=1#
#=>x^2-3x=10#
#=>x^2-3x-10=10-10#
#=>x^2-3x-10=0#
#=>(x-5)(x+2)=0#
#=>x_1=-2, x_2=5#

Related questions
See all questions in Logarithmic Models
Impact of this question
1140 views around the world
You can reuse this answer
Creative Commons License