+26 So I know that all the logs have bases of 2, so I can divide (x+1) to (x+3)
But do I divide (x+3) by (x-1)?
If I do that, then there would be a fraction in the denominator, how can I either avoid that or work that out?
The answer in the textbook says: log_2 (x+1)^2/ (x+3) (x-1). I just don't know why they multiplied instead of divided (x+3) and (x-1)
+128475 2log2(x + 1) - log2 (x + 3) - log2 (x - 1) ......we can write this as :
log2(x + 1)^2 - [ log2 (x + 3) + log2 (x - 1)] = [remember .....log a + log b = log(a * b) ]
log2 (x + 1)^2 - [ log2 ( x+ 3) (x - 1) ] = [ also...... log a - log b = log (a / b) ]
log2 [ (x + 1)^2 / [ (x + 3) (x - 1)]
