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$</span><span class="mn">54.5</span>billionin 2013to<span class="mo">$</span><span class="mn">74.6</span>billion1in 2015. Find an exponential function to model the revenue as a function of years since 2013. What is the continuous percent growth rate, per year, of revenue? Round your answer to three decimal places. </p> <p>A(t)=? </p> <p> </p> <p>The form XXX exponential XXXXXXXXX is: X(t) = X * e^(r * t) </p> <p>We XXXX XXXX XXX XXXXXX XX these XXXXXXXXX/XXXXXXXXX: P, r </p> <p>XXXX question gives initial XXXXXXXXXX. XXX <b>initial year, t<sub>0</sub> = 0</b> (XXXX corresponds XX XXXX), and the <b>XXXXXXX XXXXXXX A(t<XXX>X</sub>) = X(X) = XX.5 billion</b>. </p> <p>We use these initial conditions XX solve for the constant <b>P</b>: </p> <ol><li>XXXX in 0 XXX t : X(X) <ol><li>XX XXX the general XXXX A(t) = P * e^(r * t) </li><li> X(<b>0</b>) = X * e^(r * <b>X</b>) </li><li> X(<b>X</b>) = P * e^0 </li><li> X(<b>0</b>) = P * X </li><li> A(<b>X</b>) = X </li></ol></li><li>XX know, XXXX XXXXXXX, XXXX X(X) = 54.X billion </li><li>XX now have two XXXX XX represent X(X) <ol><li>A(X) = P </li><li>X(X) = XX.5 </li></ol></li><li>Thus X = XX.5 </li></ol> <p>Now XXXX we XXXX what P XX, all we need XX find is r. </p> <p>XX start with X(t) =<b> P</b> * e^(r * t). XXXX XXXXXX it with the value XX P XX XXXX XXXXX. </p> <p> A(t) = <b>(XX.5)</b> * e^(r * t) </p> <p> </p> <p>Now there XX another condition XXXX is given: XX XXXX, Apple made $XX.X XXXXXXX. Let t = 2 (corresponds to XXXX, X XXXXX XXXXX 2013), then A(2) = 74.6 billion. </p> <p>XXXX XXX this XX XXXXX XXX r: </p> <ol><li>XXXX in X XXX t : X(2) <ol><li>XX XXX A(t) = (54.5) * e^(r * t) </li><li> X(<b>X</b>) = (XX.X) * e^(r * <b>2</b>) </li><li> X(2) = (XX.5) * e^(XX) </li></ol></li><li>XX XXXX XXXX XXXX earlier that A(X) = XX.6 XXXXXXX </li><li>We XXX know two ways to XXXXXXXXX X(X) <ol><li>A(2) = (XX.X) * e^(2r) </li><li>A(2) = XX.X billion </li></ol></li><li>Thus 74.6 = (54.X) * e^(2r) </li><li>Solve XXX r <ol><li>XX.6 = (54.X) * e^(XX) </li><li>/XX.5 = /54.X </li><li>(74.X/54.5) = e^(XX) </li><li>XX( XX.X/54.X ) = XX( e^(2r) ) </li><li>XX( 74.X/XX.5 ) = XX </li><li>/2 = /X </li><li>XX( 74.6/54.X )/2 = r </li></ol></li><li>Thus r = ln( XX.X/XX.X )/2 = 0.15696990277 </li></ol> <p>XXXX constant<b> r,</b> represents the XXXXXXXXXX XXXXXX rate XXX year XX XXXXXXX. XX convert to XXXXXXXXXX, XX.XXXXXXXXX % XXX XXXX round XX three XXXXXXX XXXXXX: XX.7%. </p> <p>XXXX we have <b>A(t) = XX.5 e^(0.XXX t)</b>. </p> <p>XXXXXX XXX me know if you have any other questions 😊 Good XXXX! </p> <XX><li>-Connor XXXXX </li></XX> <p> </p>">
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