How do you solve #log_2(x-6)+log_2(x- 4)=log_2x#?

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Answer 1 · 2018-04-16T17:08:15

#=>x=8,3#

#log_2(x-6)+log_2(x- 4)=log_2x#

#=>log_2[(x-6)(x- 4)]=log_2x#

#=>(x-6)(x- 4)=x#

#=>x^2-10x+24-x=0#

#=>x^2-11x+24=0#

#=>x^2-8x-3x+24=0#

#=>x(x-8)-3(x-8)=0#

#=>(x-8)(x-3)=0#

#=>x=8,3#

Answer 2 · 2018-04-16T17:20:55

#color(blue)(x=8)#

By the laws of logarithms:

#log_a(b)+log)a(c)=log_a(bc)#

#log_2(x-6)+log_2(x-4)=log_2((x-6)(x-4))=log_2(x)#

If:

#log_2((x-6)(x-4))=log_2(x)#

Then:

(x-6)(x-4)=x#

Expanding and simplifying:

#x^2-10x+24=x#

#x^2-11x+24=0#

Factor:

#(x-3)(x-8)=0=>x=3 and x=8#

For #x=3#

#log_2((3)-6)=log_2(-3)#

This is not defined for real numbers:

#color(blue)(x=8)# is the only solution: