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

1 Answer
May 8, 2016

Answer:

x = 9

Explanation:

Use #log_b m + log_b n = log_b(mn)#

and inverse of #y=log_bx# is #x = b^y#

Here, # log_3 x+log_b(x-8)= log_3(x(x-8))=2#.

Inversely, # x(x-8)=3^2=9#.

So, #x^2-8x-9= (x+1)(x-9)=0#.

#x = -1 and 9#. Negative x is inadmissible for #log_3x#.