write the exponential function f(x)=-3*4^(1-x) in the form f(x)=ab^x
y = -3 * 4^1 * 4 ^-x
= -12 * 4^-x
= -12 * 1/4^x
= -12 * (1/4)^x
can you help with 2 more? i answered them but i need you to check and correct me if im wrong
1.find the zeroes of the function f(x)=log4(x+1)+log4(4x-3)
2.which of the following is the equation c^(4d+1)=7a-b written in logarithm form
Answer is C
f(x)=-12(1/4)^x
js did test
To rewrite the exponential function f(x) = -3*4^(1-x) in the form f(x) = ab^x, we need to simplify and express it with a base of b.
Let's begin by rewriting 4^(1-x):
4^(1-x) = (2^2)^(1-x) [since 4 is equal to 2^2]
= 2^(2*(1-x)) [apply the power rule for multiplying exponents]
= 2^(2-2x)
Now we can substitute this expression back into the original function, f(x):
f(x) = -3*4^(1-x)
= -3*(2^(2-2x))
To express it in the form f(x) = ab^x, we'll rearrange the terms:
= (-3*2^2)/(2^2)^x
= (-3*2^2)/(2^(2x)) [apply the power rule for dividing exponents]
= (-3*4)/(2^(2x))
= -12/(2^(2x))
Finally, we can rewrite it as:
f(x) = -12*2^(-2x)
Now, the exponential function f(x) = -3*4^(1-x) has been expressed in the form f(x) = ab^x as f(x) = -12*2^(-2x).