Question

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Answer 1

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

1. How to find which statements are true.

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

2. How to find the solution of 3y² + 5y > -2?

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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