Well, the keyword here is One of the performers insist on being the last.
So, the amount of performers that we can arrange freely is 7 performers.
Different ways we can schedule their appearance is :
7 ! = 7 x 6 x 5 x 4 x 3 x 2 x 1
= 5040
hope this helps
20 because there are many more
Hey Katie
4^2x+3 = 1
We need to solve for x
log(4^2x+3) = log (1)
Now take log of
(2x +3) * log (4) = log (1)
2x + 3 = log(1)/log(4)
2x + 3 = 0
2x = 0 - 3
2x = -3
x = -3/2
I hope that's help !
