What would the equation need to be to have zero, and one solutions?

PLEASE

Step-by-step explanation:

hello :

a^(x+2) +b =c  equivalent to : a^(x+2) = c-b

the equation need to be to have zero and one solutions? when :  a>0 and a≠0  and c > b

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.

(3x-8) or (3x+8)

step-by-step explanation: