What is x if #log_3 (2x-1) = 2 + log_3 (x-4)#?

2 Answers
Nov 26, 2015

Answer:

#x = 5#

Explanation:

We will use the following:

  • #log_a(b) - log_a(c) = log_a(b/c)#
  • #a^(log_a(b)) = b#

#log_3(2x-1) = 2 + log_3(x-4)#

#=> log_3(2x-1) - log_3(x-4) = 2#

#=>log_3((2x-1)/(x-4)) = 2#

#=> 3^(log_3((2x-1)/(x-4))) = 3^2#

#=> (2x-1)/(x-4) = 9#

#=> 2x - 1 = 9x - 36#

#=> -7x = -35#

#=> x = 5#

Nov 26, 2015

Answer:

I found: #x=5#

Explanation:

We can start writing it as:
#log_3(2x-1)-log_3(x-4)=2#
use the property of the logs: #logx-logy=log(x/y)# and write:
#log_3((2x-1)/(x-4))=2#
use the definition of log:
#log_bx=a->x=b^a#
to get:
#(2x-1)/(x-4)=3^2# rearranging:
#2x-1=9(x-4)#
#2x-9x=-36+1#
#7x=35#
#x=35/7=5#