00:01
Okay, so for this problem for part a, we're going to need ln of 1 plus x over 1 minus x to equal 2, and then x plus x to the 3 plus x to the 5 over 5, etc.
00:18
And since y equals 1 plus x over 1 minus x, we want to go ahead and multiply the denominator out.
00:26
So that way we can solve for x.
00:29
So i'm going to distribute, and then i'm going to move the x's to one side and everything else to the other side.
00:40
So i'm going to have y minus 1 equals x plus y to the x, factor out x on my right hand side.
00:49
So then now this is going to give me x equals y minus 1 over y plus 1.
00:56
So now since i have y equaling 3, then i'm going to have 3 minus 1 over 3 plus 1, which is, is going to give me 2 over 4, which is 1 .5...