(9^x) - 3 = 2*3^x
(9^x) - 3 - (2*3^x) = (2*3^x) - (2*3^x)
(9^x) - (2*3^x) - 3 = 0
(3^2)^x - 2*(3^x) - 3 = 0
3^(2x) - 2*(3^x) - 3 = 0
3^(x*2) - 2*(3^x) - 3 = 0
(3^x)^2 - 2*(3^x) - 3 = 0
z^2 - 2*z - 3 = 0 ............ let z = 3^x
(z - 3)(z + 1) = 0
If z-3 = 0, then z = 3 when we isolate z
If z = 3, and z = 3^x, then
z = 3
3^x = 3
3^x = 3^1
x = 1
which is a solutin in terms of x
If z+1 = 0 then z = -1
If z = -1 and z = 3^x, then there are NO solutions for this part of the equation
The quantity 3^x is never negative no matter what the x value is
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Answer: x = 1
Answer: C and D
Step-by-step explanation:I just did it my bad for the late response.
Answer:
The first one
Step-by-step explanation:
g(x) = ax² + bx + c
Point (0,0):
0 = a.0² + b.0 + c
c = 0
Point (2,1):
1 = a.2² + b.2 + c
4a + 2b + c = 1
But c = 0. Then:
4a + 2b = 1
Another point: (- 2, 1):
1 = a.(- 2)² + b.(- 2) + c
4a - 2b = 1
{4a + 2b = 1
{4a - 2b = 1
4a + 2b = 4a - 2b
4a - 4a = - 2b - 2b
- 4b = 0
b = 0
4a + 2b = 1
4a + 2.0 = 1
4a = 1
a = 1/4
The formula is:
g(x) = (1/4)x²
I hope I've helped you.
