sorry it's log(base6)(x+3)−log(base6)(x−5)=2
the x in front of log isn't part of it
Solve for x log(base6)(x+3)−log(base6)(x−5)=2.
I tried to factor out the log(base6) and got (x+3)^x/(x-5)=2, but I don't know how to solve that...
4 answers
Oh no, 6 is not a number, it is an operator.
You can't factor out 6
(that would be doing something like
√8 - √3
= (√)(8-3) )
You have to use your rules of logs.
6(x+3) - 6(x-5) = 2
6 ( (x+3)/(x-5) ) = 2
which by definition is
(x+3)/(x-5) = 6^2 = 36
36x - 180 = x+3
35x = 183
x = 183/35
You can't factor out 6
(that would be doing something like
√8 - √3
= (√)(8-3) )
You have to use your rules of logs.
6(x+3) - 6(x-5) = 2
6 ( (x+3)/(x-5) ) = 2
which by definition is
(x+3)/(x-5) = 6^2 = 36
36x - 180 = x+3
35x = 183
x = 183/35
Oh no, log6 is not a number, it is an operator.
You can't factor out log6
(that would be doing something like
√8 - √3
= (√)(8-3) )
You have to use your rules of logs.
log6(x+3) - log6(x-5) = 2
log6 ( (x+3)/(x-5) ) = 2
which by definition is
(x+3)/(x-5) = 6^2 = 36
36x - 180 = x+3
35x = 183
x = 183/35
You can't factor out log6
(that would be doing something like
√8 - √3
= (√)(8-3) )
You have to use your rules of logs.
log6(x+3) - log6(x-5) = 2
log6 ( (x+3)/(x-5) ) = 2
which by definition is
(x+3)/(x-5) = 6^2 = 36
36x - 180 = x+3
35x = 183
x = 183/35
thank you for your explanation :)
1 answer