Solving Exponential and Logarithmic Equations

Transcription

1 Solving Exponential and Logarithmic Equations We will now use the properties of logarithms along with the fact that exponential and logarithmic functions are inverses of each other to solve problems involving exponential and logarithmic equations Exponential Equations: To solve: 1) Try to write both sides of the equation with the same base, if possible Then set the exponents equal to each other 2) If you cannot write both sides of the equation with the same base, then take the log(using the given base or a base you ve converted to) of both sides of the equation Then solve Solve 1) 9 8x 27 x3 2) x x (3) 3 x3 log x1 log x 3 3x9 x 1 log 5 1 2x = 3x 9 x = 5 x 1 log5 log x 1 log5 log 211

2 Now, you try to solve the following problems A) 3 2x 27 B) 2 3x x 3 log 3 2 3x 2 log x 3 x = 3/2 x 3x log15 log2 log15 log You must isolate the exponential part of the equation before logging both sides of the equation For example, let s solve, log 10 2x3 log x3 17--> 10 2x > > 2x 3 ln13 ln10 --> x 111 *Note(Important): When solving Exponential and Logarithmic equations, be sure to try to keep as many decimal places in the middle steps of the problem as possible Otherwise, your final answer may be quite a bit off from what the true answer is Often, you can use the word log in your calculations without having to find out a decimal equivalent of the log in every step of the problem

3 Logarithmic Equations: To solve: 1) Try to write both sides of the equation with the same base logarithm, if possible Then set the arguments of the logarithms equal to each other 2) If you cannot write both sides of the equation with the same base logarithm, then exponentiate both sides of the equation using the given base or a base you ve converted to *OR, at some point in the problem, you could change from logarithmic notation to exponential notation which makes the problem one step shorter Then solve *3) With any equation that starts with logarithms in the equation, you MUST check your answer because you may have extraneous solutions The reason for this is because the domain of a logarithmic function is usually x > 0 or it could be x > some other number if the function has been shifted left or right Solve 1) log (2x 1) 2log x --> Before we do anything here, we must get rid of any coefficients in front of any log statement or the logarithmic properties won t work So, we bring the coefficient of 2 up as an exponent --> log (2x 1) log x 2 log (2x1) log x --> Now six both sides, 2 --> The base and log undo each other since they are inverses and we are left with 2x 1 x 2 x 2 2x 1 0 (x 1) 2 0 x 1 which checks in the original equation

4 log (2x 1) 1 2) Yes, we could six both sides, but instead let s change to exponential form --> 1 2x 1 This saved us the one step that sixing both sides would have 1 added Now, 2x 1 2x 11 x 7 2, which checks *Note(Important): Once again, as we said in the previous section, if you use Properties of Logs the base for the logs must be the SAME or somehow you can convert the base to be the same or you cannot use Properties of Logs log 100 log (x 1) 1 log 100 3) --> x > Yes, we could both sides but instead let s cut off that step by converting from log notation to exponential notation --> x 100 x 2 x 1, which checks in the original equation Notice that changing to exponential notation works only if you put all of the logs to one side of the equation leaving a constant on the other side of the equation We could have done Example 1) above the same way by setting the original equation equal to 0, but it would have been more work than simply equating the arguments of the logs Now, you try to solve the following logarithmic equations: A) C) log 8 3 log (3x 1) 2 5 log (2x 1) 3 B) D) log (5x 1) 3 log8 3log x log (x 7)

5 Answers(All answers check in the original equation): A) x = 73/1 B) x = 2 C) x = 8 D) x = 5 Sometimes you can find creative ways to get to the answers for these problems Solve, log x x x x 5 2 x 25 *Remember, you must check your answers if the problem begins with logs in it Solve log x 12 log (x 1) 1 12 log But x = - does not check in the original equation since we cannot take the log of a negative number So, the only solution here is x = 3 12 xx 1 1 log (x 2 x) 1 x 2 x 12 1 x 2 x 12 0 x 3 or x 12 Applications of Exponential and Logarithmic Equations Now that we can solve both exponential and logarithmic equations, we can model other real-life situations and solve problems in those situations Example: Plutonium 239 has a half-life of 2,110 years How long does it take for a 1 gram sample of Pu-239 to decay to 01 g?

6 This is a continuous decay problem, so our equation will of course involve e Recall the equation for continuous growth or decay, A(t) A 0 e rt For these types of problems, very similar to our direct and inverse variation problems, the rate of decay will be a constant, which we will call k So, we replace r with k in the function and it now looks like, A(t) A 0 e kt Put the numbers into the equation, 011e kt and we cannot solve it because we don t know k Just as we did with direct and inverse variation problems, we need to use some initial information given in the problem to solve for k before we can solve for t That information is Plutonium 239 has a half-life of 2,110 years We started with 1g of Plutonium 239 So, 2,110 years later, 05g will be leftover Now, the equation becomes, A(t) A 0 e kt 05 1e k2, e k2,110 Take the ln of both sides and keep a lot of decimal places for k, ln05 lne k2, ,110k k Back to the original equation: A(t) A 0 e kt --> 011e t ln both sides--> ln01 lne t t which gives us t 80,092 years Example: Suppose a bacteria culture doubles in size every hour How many hours will it take for the number of bacteria to exceed 1,000,000? N(t) 2 t Doubling means we are using the exponential function where N represents the number of bacteria at any time t after

7 the start of bacteria growth So, we need to solve the inequality, 2 t 1,000,000 Take the log base 2 of both sides: log 2 t log 1,000,000 t log1,000, hours 2 2 log2 This problem was a bit easier since we did not need to figure out any k, but problems like these could involve a growth constant as well Yes, we could have just done a log or ln to both sides of this equation and then brought the exponent t down in front of the base of 2 *Note: Radioactive elements decay continuously Now, you try to solve the following problems: A) The magnitude M of an earthquake that releases energy E, in egrs, can be modeled by the equation M 0291lnE 117 On May 22, 190, one of the most powerful earthquakes on record shook the country of Chile The earthquake had a magnitude of 95 How much energy did this earthquake release? Put your answer in scientific notation B) Determine how long it will take for a 50mg sample of Chromium-51, which has a half-life of about 28 days, to decay to 200mg *Keep at least 5 decimal places for k A(t) A 0 e kt Answers: A) E 2702 x ergs B) k ; t 7 years *Hmwk: Do the first two problems below and do odd, 3- all, 8-70 all shown below Also, read Section 5 in your textbook and do p339(29-35 odd) and p30(71, 73)

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