How do you solve #(log_2(2x-3) / log_2(x)) - log_x(x+6) + (1/log_ (x+2)(x)) = 1#?

1 Answer
ali ergin
Mar 6, 2016

#x^3+8x^2+10x+3=0#

Explanation:

#(log_2(2x-3)/log_2 x)-log_x(x+6)+(1/log_(x+2) x)=1#
#log_x(x+6)=(log_2 (x+6))/(log_2x)#
#log_(x+2) x=(log_2 x)/(log_2(x+2))#
#(log_2(2x-3)/log_2 x)-log_2(x+6)/(log_2 x)+log_2(x+2)/log_2 x=1#
#(log_2(2x-3)-log_2(x+6)+log_2(x+2))/(log_2 x)=1#
#log_2(2x-3)-color(red)(log_2(x+6)+log_2(x+2))=log_2 x#
#color(green)(log_2(2x-3)-log_2 (x+6)*(x+2))=log_2 x#
#log_2 ((2x-3)/((x+6)(x+2)))=log_2 x#
#(2x-3)/((x+6)(x+2))=x#
#x(x^2+2x+6x+12)=2x-3#
#x^3+2x^2+6x^2+12x-2x+3=0#
#x^3+8x^2+10x+3=0#

Related questions
See all questions in Logarithmic Models
Impact of this question
1299 views around the world
You can reuse this answer
Creative Commons License