Question

write exponential function f(x)=-3*4^(1-x) in the form f(x)=ab^x
A. f(x)=-12*(-4)^x
B. f(x)=-12*(-1/12)^x
C. f(x)=-12*(1/4)^x
D. f(x)=-3*(1/4)^x

Answer 1

find the zero(es) of the function f(x)=log_3(x-1)+log_3(2x+3)
A. x=2 and x=-1
B. x=1 and x=-2/3
C.-1+ sqr 33/4
D.-1+- sqr 33/4

Answer 2

To write the exponential function f(x)=-3*4^(1-x) in the form f(x)=ab^x, we can rewrite the expression by factoring -3 and expressing 4 as a power of another base.

f(x) = -3*4^(1-x)

= -3*2^2(1-x)

= -3*(2^(2(1-x)))

= -3*(2^(2-2x))

Thus, the exponential function f(x) can be rewritten as f(x) = -3*(1/4)^x.

Therefore, the correct answer is D.

Answer 3

To find the zero(es) of the function f(x) = log_3(x-1) + log_3(2x+3), we need to find the value(s) of x that make the function equal to zero.

First, let's rewrite the equation in exponential form:

log_3(x-1) + log_3(2x+3) = 0

Using the property of logarithms that states that log_a(x) + log_a(y) = log_a(xy), we can combine the two logarithms:

log_3((x-1)(2x+3)) = 0

Now, let's convert this logarithmic equation into an exponential equation:

3^0 = (x-1)(2x+3)

1 = (x-1)(2x+3)

Expanding the right side of the equation:

1 = 2x^2 + x - 3

Rearranging the equation to the standard quadratic form:

2x^2 + x - 4 = 0

To find the zeros, we can use factoring, completing the square, or the quadratic formula. In this case, we'll use factoring:

(2x + 4)(x - 1) = 0

Setting each factor equal to zero:

2x + 4 = 0 or x - 1 = 0

Solving the first equation:

2x = -4

x = -2

Solving the second equation:

x = 1

Therefore, the zero(es) of the function f(x) = log_3(x-1) + log_3(2x+3) are x = -2 and x = 1.

So the correct answer is A. x = 2 and x = -1.