Applications of Exponential and Logarithmic Functions

To find the interval in cents for a frequency that changes from 420Hz to 415Hz, we will substitute a=420 and b=415 in the given formula. Then we will evaluate the right-hand side. To do so, we will use the Change of Base Formula to be able to use a calculator.

If a frequency changes from 420Hz to 415Hz, the interval is about 20.73 cents.

We know that the final frequency is 218Hz and that the interval is 60 cents. To find the initial frequency, we will substitute b=218 and n=60 into the given formula and solve for the initial frequency a.

Now, to isolate a, we will use the definition of a logarithm. Definition y=log_b x ⇔ b^y= x [1em] Equation 1/20=log_2 a/218 ⇔ 2^(120)= a/218 Finally, let's solve this equation for a.

If the interval is 60 cents and the final frequency is 218Hz, then the beginning frequency is about 225.69Hz.

$x$	$y$
$1$	$3$
$2$	$\approx - 0.47$
$3$	$\approx - 2.49$
$4$	$\approx - 3.93$
$5$	$\approx - 5.05$

$x$	$y$
$1$	$1 k$
$2$	$2 k$
$3$	$4 k$
$4$	$8 k$
$5$	$16 k$

$x$	$40 e^{- 0.154 t}$	$y$
$0$	$40 e^{- 0.154 (0)}$	$40$
$2$	$40 e^{- 0.154 (2)}$	$\approx 29.4$
$4$	$40 e^{- 0.154 (4)}$	$\approx 21.6$
$6$	$40 e^{- 0.154 (6)}$	$\approx 15.9$
$8$	$40 e^{- 0.154 (8)}$	$\approx 11.7$
$10$	$40 e^{- 0.154 (10)}$	$\approx 8.6$

$x$	$x ln 2$	$ln y = x ln 2$
$- 2$	$- 2 ln 2$	$\approx - 1.39$
$- 1$	$- 1 ln 2$	$\approx - 0.69$
$0$	$0 ln 2$	$0$
$1$	$1 ln 2$	$\approx 0.69$
$2$	$2 ln 2$	$\approx 1.39$

$x$	$y$	$ln y$
$20$	$12$	$ln 12 \approx 2.48$
$30$	$15$	$ln 15 \approx 2.71$
$40$	$25$	$ln 25 \approx 3.22$
$50$	$40$	$ln 40 \approx 3.69$
$60$	$100$	$ln 100 \approx 4.61$

Applications of Exponential and Logarithmic Functions

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Tables, Logarithmic Models, and Exponential Models

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Solution

Carbon $14$ in Vegetables

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Solution

Uranium $238$

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Solution

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Solution

Relation Between Exponential and Linear Models

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Answer

Hint

Solution

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Applications of Exponential and Logarithmic Functions

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