Asked by Ann

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...
Answered by Ann
sorry it's log(base6)(x+3)−log(base6)(x−5)=2

the x in front of log isn't part of it

Answered by Reiny
Oh no, <sub>6</sub> is not a number, it is an operator.

You can't factor out <sub>6</sub>
(that would be doing something like
√8 - √3
= (√)(8-3) )

You have to use your rules of logs.

<sub>6</sub>(x+3) - <sub>6</sub>(x-5) = 2
<sub>6</sub> ( (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
Answered by Reiny
Oh no, log<sub>6</sub> is not a number, it is an operator.

You can't factor out log<sub>6</sub>
(that would be doing something like
√8 - √3
= (√)(8-3) )

You have to use your rules of logs.

log<sub>6</sub>(x+3) - log<sub>6</sub>(x-5) = 2
log<sub>6</sub> ( (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
Answered by Ann
thank you for your explanation :)
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