A population of 50 flies is expected to double every week, leading to a function of the x

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1 4 Setion 4.3 Logarithmi Funtions population of 50 flies is epeted to doule every week, leading to a funtion of the form f ( ) 50(), where represents the numer of weeks that have passed. When will this population reah 500? Trying to solve this prolem leads to: () Dividing oth sides y 50 to isolate the eponential 10 While we have set up eponential models and used them to make preditions, you may have notied that solving eponential equations has not yet een mentioned. The reason is simple: none of the algerai tools disussed so far are suffiient to solve eponential equations. Consider the equation 10 aove. We know that 3 8 and 4 16, so it is lear that must e some value etween 3 and 4 sine g ( ) is inreasing. We ould use tehnoy to reate a tale of values or graph to etter estimate the solution. From the graph, we ould etter estimate the solution to e around 3.3. This result is still fairly unsatisfatory, and sine the eponential funtion is one-to-one, it would e great to have an inverse funtion. None of the funtions we have already disussed would serve as an inverse funtion and so we must introdue a new funtion, named as the inverse of an eponential funtion. Sine eponential funtions have different ases, we will define orresponding arithms of different ases as well. Logarithm The arithm (ase ) funtion, written ( ) funtion (ase ),., is the inverse of the eponential Sine the arithm and eponential are inverses, it follows that: Properties of Logs: Inverse Properties ( )

2 Setion 4.3 Logarithmi Funtions 43 Reall also from the definition of an inverse funtion that if pplying this to the eponential and arithmi funtions: 1 f ( a), then f ( ) a. Logarithm Equivalent to an Eponential The statement a is equivalent to the statement ( ) a. a lternatively, we ould show this y starting with the eponential funtion, then a taking the ase of oth sides, giving ( ). Using the inverse property of s we see that ( ) a. Sine is a funtion, it is most orretly written as ( ), using parentheses to denote funtion evaluation, just as we would with f(). However, when the input is a single variale or numer, it is ommon to see the parentheses dropped and the epression written as. Eample 1 Write these eponential equations as arithmi equations: is equivalent to (8) is equivalent to 5 (5) 10 1 is equivalent to Eample Write these arithmi equations as eponential equations: 1 6 ( 6) 3 ( 9) ( 6) 1 6 is equivalent to 6 1 / 6 3 is equivalent to 3 9 ( 9) Try it Now Write the eponential equation 4 16 as a arithmi equation.

3 44 By estalishing the relationship etween eponential and arithmi funtions, we an now solve asi arithmi and eponential equations y rewriting. Eample 3 Solve 4 ( ) for. By rewriting this epression as an eponential, 4, so 16 Eample 4 Solve 10 for. By rewriting this epression as a arithm, we get (10) While this does define a solution, and an eat solution at that, you may find it somewhat unsatisfying sine it is diffiult to ompare this epression to the deimal estimate we made earlier. lso, giving an eat epression for a solution is not always useful often we really need a deimal approimation to the solution. Lukily, this is a task alulators and omputers are quite adept at. Unlukily for us, most alulators and omputers will only evaluate arithms of two ases. Happily, this ends up not eing a prolem, as we ll see riefly. Common and Natural Logarithms The ommon is the arithm with ase 10, and is typially written ( ). The natural is the arithm with ase e, and is typially written ln( ). Eample 5 Evaluate ( 1000) using the definition of the ommon. To evaluate ( 1000), we an say (1000), then rewrite into eponential form using the ommon ase of From this, we might reognize that 1000 is the ue of 10, so 3. We also an use the inverse property of s to 3 write ( 10 ) 3 10 Values of the ommon numer numer as (numer) eponential

4 Setion 4.3 Logarithmi Funtions 45 Try it Now. Evaluate ( ). Eample 6 Evaluate ln ( e ). We an rewrite ( e ) ln as ln( e 1/ ) property for s: ln( 1/ ) ( e 1/ ). Sine ln is a ase e, we an use the inverse 1 e e. Eample 7 Evaluate (500) using your alulator or omputer. Using a omputer, we an evaluate ( 500) To utilize the ommon or natural arithm funtions to evaluate epressions like (10), we need to estalish some additional properties. Properties of Logs: Eponent Property r r ( ) ( ) To show why this is true, we offer a proof. Sine the arithmi and eponential funtions are inverses, r So ( ) r p Utilizing the eponential rule that states ( ) q pq, r r r ( ) r r So then ( ) ( ) gain utilizing the inverse property on the right side yields the result r r ( ). Eample using the eponent property for s. Rewrite ( ) Sine 5 5, ( ) ( ) 5 3 3

5 46 Eample 9 Rewrite 4ln( ) using the eponent property for s. 4 Using the property in reverse, 4ln( ) ln( ) Try it Now 1 3. Rewrite using the eponent property for s: ln. The eponent property allows us to find a method for hanging the ase of a arithmi epression. Properties of Logs: Change of Base ( ) ( ) ( ) Proof: Let ( ). Rewriting as an eponential gives. Taking the ase of oth sides of this equation gives Now utilizing the eponent property for s on the left side, Dividing, we otain or replaing our epression for, With this hange of ase formula, we an finally find a good deimal approimation to our question from the eginning of the setion. Eample 10 Evaluate (10) using the hange of ase formula. ording to the hange of ase formula, we an rewrite the ase as a arithm of any other ase. Sine our alulators an evaluate the natural, we might hoose to use the natural arithm, whih is the ase e: e 10 ln10 10 e ln Using our alulators to evaluate this,

6 Setion 4.3 Logarithmi Funtions 47 ln10 ln This finally allows us to answer our original question the population of flies we disussed at the eginning of the setion will take 3.3 weeks to grow to 500. Eample 11 Evaluate 5 (100) using the hange of ase formula. We an rewrite this epression using any other ase. If our alulators are ale to evaluate the ommon arithm, we ould rewrite using the ommon, ase (100) While we were ale to solve the asi eponential equation 10 y rewriting in arithmi form and then using the hange of ase formula to evaluate the arithm, the proof of the hange of ase formula illuminates an alternative approah to solving eponential equations. Solving eponential equations: 1. Isolate the eponential epressions when possile. Take the arithm of oth sides 3. Utilize the eponent property for arithms to pull the variale out of the eponent 4. Use algera to solve for the variale. Eample 1 Solve 10 for. Using this alternative approah, rather than rewrite this eponential into arithmi form, we will take the arithm of oth sides of the equation. Sine we often wish to evaluate the result to a deimal answer, we will usually utilize either the ommon or natural. For this eample, we ll use the natural : ln ( ) ln(10) Utilizing the eponent property for s, ln ( ) ln(10) Now dividing y ln(), ln(10).861 ln ( ) Notie that this result mathes the result we found using the hange of ase formula.

7 48 Eample 13 In the first setion, we predited the population (in illions) of India t years after 008 t y using the funtion f ( t) 1.14( ). If the population ontinues following this trend, when will the population reah illion? We need to solve for the t so that f(t) t 1.14(1.0134) Divide y 1.14 to isolate the eponential epression t Take the arithm of oth sides of the equation t ln ln( ) 1.14 pply the eponent property on the right side ln t ln( ) 1.14 Divide oth sides y ln(1.0134) ln 1.14 t 4.3 years ln ( ) If this growth rate ontinues, the model predits the population of India will reah illion aout 4 years after 008, or approimately in the year 050. Try it Now 4. Solve 5 (0.93) 10. In addition to solving eponential equations, arithmi epressions are ommon in many physial situations. Eample 14 In hemistry, ph is a measure of the aidity or asiity of a liquid. The ph is related to the onentration of hydrogen ions, [H + ], measured in moles per liter, y the equation ph H +. ( ) If a liquid has onentration of moles per lier, determine the ph. Determine the hydrogen ion onentration of a liquid with ph of 7. To answer the first question, we evaluate the epression ( ). While we ould use our alulators for this, we do not really need them here, sine we an use the inverse property of s: ( 4) ( ) ( ) 4

8 Setion 4.3 Logarithmi Funtions 49 To answer the seond question, we need to solve the equation 7 ( H + ). Begin y isolating the arithm on one side of the equation y multiplying oth sides y -1: ( ) 7 H + Rewriting into eponential form yields the answer 7 H moles per liter. Logarithms also provide us a mehanism for finding ontinuous growth models for eponential growth given two data points. Eample 15 population grows from 100 to 130 in weeks. Find the ontinuous growth rate. Measuring t in weeks, we are looking for an equation P() 130. Using the first pair of values, r ae, so a 100. P rt ( t) ae so that P(0) 100 and Using the seond pair of values, e r Divide y r e Take the natural of oth sides 100 r ln(1.3) ln( e ) Use the inverse property of s ln(1.3) r ln(1.3) r This population is growing at a ontinuous rate of 13.1% per week. In general, we an relate the standard form of an eponential with the ontinuous growth form y noting (using k to represent the ontinuous growth rate to avoid the onfusion of using r in two different ways in the same formula): k a ( 1+ r) ae k ( 1+ r ) e k 1 + r e Using this, we see that it is always possile to onvert from the ontinuous growth form of an eponential to the standard form and vie versa. Rememer that the ontinuous growth rate k represents the nominal growth rate efore aounting for the effets of ontinuous ompounding, while r represents the atual perent inrease in one time unit (one week, one year, et.).

9 50 Eample 16 ompany s sales an e modeled y the funtion years. Find the annual growth rate. S 0.1t ( t) 5000e, with t measured in k 0.1 Noting that 1 + r e, then r e , so the annual growth rate is 1.75%. t The sales funtion ould also e written in the form S ( t) 5000( ). Important Topis of this Setion The Logarithmi funtion as the inverse of the eponential funtion Writing arithmi & eponential epressions Properties of s Inverse properties Eponential properties Change of ase Common Natural Solving eponential equations Try it Now nswers ln( ) ln() ln(0.93) 1. ( ) 4