If f(x) = x + 1, and g(x) = 2x,
then f(g(x))= __x+__

f (g(x) = f (2x) = 2x + 1 (∵ f (x) = x + 1)

2x+1=0 is the answer

(g(x) = f (2x) = 2x + 1 (∵ f (x) = x + 1)-

26?

step-by-step explanation:

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.