How do you solve #log_6x+log_6(x-9)=2#?

Question

10555 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-24T03:31:30

Use the log property #log_a(n) + log_a(m) = log_a(n xx m)#:

#=>log_6(x(x - 9)) = 2#

#=>log_6(x^2 - 9x) = 2#

#=>x^2 - 9x = 6^2#

#=>x^2 - 9x = 36#

#=>x^2 - 9x - 36 = 0#

#=>(x - 12)(x + 3) = 0#

#=>x = 12 and -3#

Checking in the original equation, we find that only #x = 12# works. Hence, the solution set is #{12}#.

Hopefully this helps!