Re write it in so that to have one single entity on the left & same on the right
log(x-1)-log5 = log[(x-1)/5]
and log(x+1)-logx = log[(x+1)/x]
So:
log[(x-1)/5] = log[(x+1)/x] ==> (x-1)/5 = (x+1)/x
Now you can solve it: x(x-1)=5(x+1)
==> x² - 6x -5 =0
Solving this quadratic equation gives x'=3+√14 & x" =3-√14
Answer:
I think it might be a I don't know for sure though I just need some more points so I can ask a question myself sorry if this didn't help
Answer:
our answer will be in quadrant 1 only.
Step-by-step explanation:
From the
cos−1(1) we have the side adjacent to ∠θ=1 and the side hypotenuse=2 . So this is a 30∘−60∘−90∘ triangle and θ=60∘therefore the opposite to ∠θ=√3.
Note that from the restrictions for the range of inverse circular functions cos−1x is restricted to quadrants 1 and 2 and since the argument is positive 12 our answer will be in quadrant 1 only.
Answer: It says sss is wrong which one is sss
Step-by-step explanation:
Answer:
LCD=24..hope it helps you