How do you solve #log_b(x-1)+log_b3=log_bx#?

Question

3066 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-14T01:37:59

I'm assuming you want to solve for x...

#log_b(3) = log_b(x) - log_b(x - 1)#

#log_b(3) = log_b((x)/(x - 1))#

Since we're in equal bases we can eliminate:

#3 = x/(x - 1)#

#3(x - 1) = x#

#3x - 3 = x#

#2x = 3#

#x = 3/2#

Hopefully this helps!