M1410304

# M1410304 - 3.27 SECTIONS 3.4 AND 3.5: EXPONENTIAL AND LOG...

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3.27 SECTIONS 3.4 AND 3.5: EXPONENTIAL AND LOG EQUATIONS AND MODELS PART A: ONE-TO-ONE PROPERTIES (Assume that b is nice.) The b x and log b x functions are one-to-one. (Their graphs pass the HLT.) Therefore, 1) b M = b N ( ) M = N ( ) , and 2) log b M = log b N ( ) M = N ( ) , where M and N are positive in value Informally: We can insert or delete the same base or the same log on both sides in order to obtain an equivalent equation, provided that we only take logs of positive values. Technical Note: The directions are immediate in both 1) and 2). The directions require the one-to-one properties of the two functions. PART B: SOLVING EXPONENTIAL EQUATIONS Exponential equations contain variable powers of constant bases, such as 3 x + 1 . Example Solve 2 x 10 = 0 . Approximate your answer to 4 decimal places. Solution We can easily isolate the basic exponential expression, 2 x , on one side. 2 x 10 = 0 2 x = 10 We’d like to bring variable exponents “down to earth.” We can do so by inserting logs.

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3.28 Method 1 Take log 2 of both sides. log 2 2 x = log 2 10 We can then exploit the Inverse Properties. x = log 2 10 If we would like a decimal approximation for x , we need to use the Change of Base Formula. x = ln10 ln2 3.3219 Method 2 (more common) Since we usually prefer calculating with ln instead of log 2 , we may prefer taking ln of both sides of 2 x = 10 . 2 x = 10 x = ln10 x Just a number = ln10 Power / "Smackdown" Rule for Logs ( ) x = ln10 x 3.3219 The solution set is: about 3.3219 { } Warning: Some instructors prefer a solution set as your final answer.
3.29 Example Solve 5 3 x 1 = 125 . Solution Although we could take log 5 or ln of both sides, it may be easier to recognize immediately that 125 = 5 3 . 5 3 x 1 = 5 3 We can then delete the 5 base on both sides. In general, we can equate exponents on the same base; i.e., b M = b N ( ) M = N ( ) . 3 x 1 = 3 3 x = 4 x = 4 3 The solution set is: 4 3

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3.30 PART C: EXPONENTIAL MODELS; MORE ON SOLVING EQUATIONS The Malthusian Population Growth Model P = P 0 e rt , where P = population at time t P 0 = initial population (i.e., population at time t = 0) r = a parameter indicating population growth (as a decimal) See Notes 3.33. ( ) Notice the similarities between this formula and the formula for continuous compound interest. Be aware that the P 0 here takes on the role of the P from the interest formulas. If r < 0 , then we actually have a decay model. Example Use the Malthusian model P = P 0 e 0.0138 t to model the population of Earth, where t = 0 corresponds to January 1, 2000. If the population of Earth on January 1, 2000 was about 6.083 billion, in what year will the population reach 20 billion?
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## This note was uploaded on 09/08/2011 for the course MATH 141 taught by Professor Staff during the Fall '11 term at Mesa CC.

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M1410304 - 3.27 SECTIONS 3.4 AND 3.5: EXPONENTIAL AND LOG...

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