Seth was studying a population of fruit flies. The population, P, after t days of the study is modeled by the function P(t)=10e^(0.5 t). What function could sech use to find D, the day the number of fruit flies reached a given value, p ? A. D=(1)/(10)ln(2p) B. D=(1)/(5)ln(b) c. D=(1)/(2)ln(10 p) D. D=2ln((p)/( 10)) ED=10 ln((p)/(2))

Set Population Equation: Seth has the function P(t) = 10e^(0.5t) which models the population of fruit flies after t days. To find the day D when the population reaches a given value p, we need to solve for t in terms of p.Isolate Exponential Term: First, we set P(t) equal to p to solve for t. p = 10e^(0.5t)Take Natural Logarithm: Next, we divide both sides of the equation by 10 to isolate the exponential term. (p/10) = e^(0.5t)Simplify Exponential Term: Now, we take the natural logarithm (ln) of both sides to solve for t. The natural logarithm is the inverse function of the exponential function, which allows us to solve for the exponent. ln(p/10) = ln(e^(0.5t))Solve for t: Using the property of logarithms that ln(e^x) = x, we can simplify the right side of the equation. ln(p/10) = 0.5tExpress as Day D: Finally, we solve for t by dividing both sides of the equation by 0.5. t = (2 * ln(p/10))We can now express t as D, the day the number of fruit flies reached the given value p. D = 2ln(p/10)

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Seth was studying a population of fruit flies. The population,
P, after
t days of the study is modeled by the function
P(t)=10e^(0.5 t).
What function could sech use to find
D, the day the number of fruit flies reached a given value,
p ?
A.
D=(1)/(10)ln(2p)
B.
D=(1)/(5)ln(b)
c.
D=(1)/(2)ln(10 p)
D.
D=2ln((p)/( 10))

ED=10 ln((p)/(2))

Seth was studying a population of fruit flies. The population, $P$ , after $t$ days of the study is modeled by the function $P(t)=10 e^{0.5 t}$ . $\newline$ What function could sech use to find $D$ , the day the number of fruit flies reached a given value, $p$ ? $\newline$ A. $D=\frac{1}{10} \ln (2 p)$ $\newline$ B. $D=\frac{1}{5} \ln (b)$ $\newline$ c. $D=\frac{1}{2} \ln (10 p)$ $\newline$ D. $D=2 \ln \left(\frac{p}{10}\right)$ $\newline$ $E. D=10 \ln \left(\frac{p}{2}\right)$

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Q. Seth was studying a population of fruit flies. The population,

P

, after

t

days of the study is modeled by the function

P(t)=10 e^{0.5 t}

\newline

What function could sech use to find

D

, the day the number of fruit flies reached a given value,

p

\newline

D=\frac{1}{10} \ln (2 p)

\newline

D=\frac{1}{5} \ln (b)

\newline

D=\frac{1}{2} \ln (10 p)

\newline

D=2 \ln \left(\frac{p}{10}\right)

\newline

E. D=10 \ln \left(\frac{p}{2}\right)

Set Population Equation: Seth has the function $P(t) = 10e^{0.5t}$ which models the population of fruit flies after $t$ days. To find the day $D$ when the population reaches a given value $p$ , we need to solve for $t$ in terms of $p$ .
Isolate Exponential Term: First, we set $P(t)$ equal to $p$ to solve for $t$ . $\newline$ $p = 10e^{(0.5t)}$
Take Natural Logarithm: Next, we divide both sides of the equation by $10$ to isolate the exponential term. $\newline$ $\frac{p}{10} = e^{0.5t}$
Simplify Exponential Term: Now, we take the natural logarithm ( $\ln$ ) of both sides to solve for $t$ . The natural logarithm is the inverse function of the exponential function, which allows us to solve for the exponent. $\newline$ $\ln(\frac{p}{10}) = \ln(e^{0.5t})$
Solve for $t$ : Using the property of logarithms that $\ln(e^x) = x$ , we can simplify the right side of the equation. $\newline$ $\ln(\frac{p}{10}) = 0.5t$
Express as Day D: Finally, we solve for $t$ by dividing both sides of the equation by $0.5$ . $t = \left(2 \cdot \ln\left(\frac{p}{10}\right)\right)$
Express as Day D: Finally, we solve for $t$ by dividing both sides of the equation by $0.5$ . $t = \frac{2 \cdot \ln(\frac{p}{10})}{0.5}$ We can now express $t$ as $D$ , the day the number of fruit flies reached the given value $p$ . $D = 2\ln(\frac{p}{10})$