Anonymous
Anonymous asked in Science & MathematicsMathematics · 5 years ago

exponentials and logarithms?

f(x)=Axe^-kx

find the value of f(4).

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• Huh
Lv 6
5 years ago

Easy! Just let x = 4, as instructed.

Equation 1: f(4) = A(4) * e^(-4 * k)

No amount of guessing will tell me what A or k is.

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f(1) = 5, okay then:

Equation 2: 5 = A * (1) * e^(-k)

f(2) = 7. Okay, then:

Equation 3: 7 = A * (2) * e^(-2k)

Divide equation 2 over equation 3 to eliminate A to 1.

5 / 7= [A * (1) * e^(-k)] / [A * (2) * e^(-2k)]

5 / 7 = ((e^(-k)) / ((2 * e^(-2k)

Multiply out the 2, it's in the way of my reasoning.

10 / 7 = [e^(-1 * k)] / [e^(-2 * k)

(10 / 7) = { [e^(k)]^(-1) } / { [e^(k)]^(-2) }

Let e^k = c. I chose c because it's my jam.

(10 / 7) = [c^(-1)] / [c^(-2)]

(10 / 7) = c^[-1 - (-2)] >>> you do understand this, right?

(10 / 7) = c^(1) = c

Undo the substitution:

10 / 7 = e^k

Take the natural logarithm of both sides of the equation because Euler's number.

ln(10 / 7) = k * ln(e), as ln(e) = 1.

ln(10 / 7) = k

ln(10) - ln(7) = k

0.3567 approx. = k

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Find A from either equation 2 or equation 3:

5 = A * e^(-k)

Divide both sides by A:

5 / A = e^(-1 * k)

5 / A = [(e)^k]^(-1)

5 / A = 1 / e^(k)

Cross multiply:

5 * e^(k) = A(1)

And what was k?

5 * e^[ln(10 / 7)] = A

That e raised to the log base e, totally cancels that logarithm bro, or sis.

5 * (10 / 7) = A

50 / 7 = A

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Solution set: A = 50 / 7, and k = ln(10 / 7)

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Finally, back to the main equation for the main event!!!!!!!!!!!!!!!!!!!! The moment that you've all been waiting for.

f(4) = A(4) * e^(-4 * k)

f(4) = (50 / 7) * (4) * e^[-4 * ln(10 / 7)]

Kick the -4 into the natural logarithm as a power :

f(4) = (200 / 7) * e^{ [ln(10 / 7)^(-4)] }

The natural logarithm gets destroyed:

f(4) = (200 / 7) * (10 / 7)^(-4)

f(4) = (200 / 7) * (7 / 10)^4

f(4) = (200 / 7) * (7^4 / 10^4)

f(4) = (200 / 7) * (2401 / 10000)

f(4) = (2401 / 7) * (200 / 10000)

f(4) = 343 * (1 / 50)

f(4) = 343 / 50 = 6.86