How do you solve #log_3x+log_3(x-8)=2#?

Question

6422 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-31T13:54:27

Thus, x=9 is the valid solution

#log_3 x+log_3 (x-8)=2#

#logm+logn=logmn#

#log_3 x+log_3 (x-8)=log_3 x(x-8)#

#log_3 x(x-8)=2#

#x(x-8)=3^2#

#x(x-8)=9#

#x^2-8x=9#

#x^2-8x-9=0#

#x^2-9x+1x-9=0#

#x(x-9)+1(x-9)=0#

#(x+1)(x-9)=0#

#x=-1, x=9#

#x=-1,9#

Here, x needs to be a positive number .

Thus, x=9 is the valid solution