how to write these log expressions as one expression?

Question

equation:
1/3 log (base 5) of (x+1) + 1/3 log (base 5) of (x-1)

do i move the 1/3 first then multiply (x+4) and (x-1) or vice versa?

equation:
log (base 3) of (x) + log (base 3) of (sqrt x+3) - 1/3 log (base 3) of (x-4)

thanks for the help.

Answer 1

For the first part you are on the right track.
It doesn't really matter the order.
You can remove the 1/3 first and say the equation is eqivalent to:
1/3 (log(x+1)+log(x-1)) = 1/3 (log{ x+1)*(x-1) })
and then even take the 1/3 inside the log function:
log{ [ (x+1)*(x-1)]^1/3 }

Or you can do it in the other order and get the same answer (I left out the (base 5) but it carries through.

For the second 1, I'll leave the (base 3) out but pretend it's there. For this you should take the 1/3 inside the log FIRST:

logx + log(sqrtx+3) - 1/3 log(x-4) =
logx + log(sqrt x + 3) - log {(x-4)^1/3 } =
log( x * (sqrt x + 3) / [(x-4)^(1/3)] )

Answer 2

I am not writing the base, since they are same in one problem. Assume the base same when you see a log operation.
Part 1//1/3 log(x+1) + 1/3 log(x-1) = 1/3 {log(x+1) + log(x-1)} = 1/3 {log(x+1).(x-1)} = 1/3 log(x^2 - 1^2)} = log(x^2 - 1)^1/3 or you may write 1/3log(x^2 - 1)
Part 2// log(x) + log(sqrt x + 3) - 1/3 log(x-4) = log{x.(sqrt x +3).(cube root (x-4)}

Answer 3

1.log(x+1)^1/3(x-1)^1/3
=log [(x+1)(x-1)]^1/3
=log(x^2-1)^1/3

2.logx(rtx+3)/(x-4)^1/3