1. Given the functions f(x) = log₂(5x) and g(x) = 5ˣ - 2 only statement c is true
2. The solution of 3y² + 5y > -2 is a. x < –1 or x is greater than negative two thirds
<h3>1. How to find which statements are true. </h3>
Statement a
Since f(x) = log₂(5x) which is a logarithm function is undefined for (-∞, 0) and defined for (0, +∞) and g(x) = 5ˣ - 2 which is an exponential function is defined for (-∞, +∞).
Also, since f(x) is decreasing on the interval (0, 1/5) while g(x) decreases on the interval (-∞, 0). So, they have do not have a common interval on (0, 1).
So, statement a. Both f(x) and g(x) decrease on the interval of (–∞, 1).
is false
Statement b
Since f(x) = log₂(5x) which is a logarithm function is defined for (0, +∞) and g(x) = 5ˣ - 2 which is an exponential function is defined for (-∞, +∞).
So, the statement b Both f(x) and g(x) have the same domain of (0, ∞) is false
Statement c
Since f(x) = log₂(5x) which is a logarithm function has a range of (0, +∞). and g(x) = 5ˣ - 2 which is an exponential function is has a range of (-2, +∞).
So, they have a common interval of (0, +∞).
So, the statement c. Both f(x) and g(x) have a common range on the interval (–2, ∞) is true
Statement d
To find the x-intercept of f(x), we equate f(x) to zero.
So, f(x) = log₂(5x)
0 = log₂(5x)
2⁰ = 5x
1 = 5x
x = 1/5
To find the x-intercept of g(x), we equate g(x) to zero.
g(x) = 5ˣ - 2
0 = 5ˣ - 2
2 = 5ˣ
x = ㏒₅2
Since the x-intercept of f(x) = 1/5 and the x- intercept of g(x) = ㏒₅2. So, they do not have a common x - intercept.
So, the statement d. Both f(x) and g(x) have the same x-intercept of (1, 0) is false.
So, only statement c is true
<h3>2. How to find the solution of 3y² + 5y > -2?</h3>
3y² + 5y > -2
3y² + 5y + 2 > 0
3y² + 3y + 2y + 2 > 0
3y(y + 1) + 2(y + 1) > 0
(3y + 2)(y + 1) > 0
So, the boundary values are at
(3y + 2)(y + 1) = 0
(3y + 2) = 0 or (y + 1) = 0
y = -2/3 or y = -1
So, we require (3y + 2)(y + 1) > 0
For y < -1 say -2, (3y + 2)(y + 1) = (3(-2) + 2)((-2) + 1)
= (-6 + 2)(-2 + 1)
= -4(-1)
= 4 > 0
For -1 < y < -2/3 say -1/3, (3y + 2)(y + 1) = (3(-1/3) + 2)((-1/3) + 1)
= (-1 + 2)(-1 +3)/2
= 1(-2/2)
= -1 < 0
For y > -2/3 say 0, (3y + 2)(y + 1) = (3(0) + 2)((0) + 1)
= (0 + 2)(0 + 1)
= 2(1)
= 2 > 0
So, for (3y + 2)(y + 1) > 0, y < -1 or y > -2/3
So, the solution of 3y² + 5y > -2 is a. x < –1 or x is greater than negative two thirds
Learn more about logarithm and exponential function here:
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