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For the exponential function A = 0.46*74.4x, what is the decay factor?

User Filhit
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1 Answer

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Answer: The decay factor is approximately 0.9984.

Explanation:

In an exponential function of the form A = Ab^x, the decay factor b is the constant factor by which the function decays or grows over each unit of time or input change. It is always a value between 0 and 1 in the case of exponential decay.

To find the decay factor in the given exponential function A = 0.46*74.4^x, we need to rewrite the function in the form A = Ab^x. We can do this by dividing both sides of the equation by the initial value of A:

A/A = (0.46*74.4^x)/A

Simplifying this expression gives us:

1 = 0.46*(74.4^x)/A

Dividing both sides by 0.46 and rearranging, we get:

(74.4^x)/A = 1/0.46

Simplifying further gives us:

(74.4^x)/A = 2.1739

Now, we can solve for the decay factor b by taking the logarithm of both sides of the equation:

log(b) = log(74.4)/ln(10)

Simplifying this expression using a calculator gives us:

log(b) = -0.0016

Taking the antilogarithm of both sides, we get:

b = 10^(-0.0016)

Approximating this value to four decimal places gives us:

b ≈ 0.9984

Therefore, the decay factor for the given exponential function is approximately 0.9984.

User Sergey Mashkov
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