Haberman MTH 111c Section II: Eponential and Logarithmic Functions Module 6: Solving Eponential Equations and More EXAMPLE: Solve the equation 10 = 100 for. Obtain an eact solution. This equation is so easy to solve that most of us can do it in our heads, without showing 2 any work. (Clearly, = 2 since 10 = 100.) But it might be helpful to notice what work would have helped us solve this equation, so that we can learn how to solve eponential equations that we can t do in our heads. If it is possible, the easiest way to solve eponential equations is to write both sides of the equation as a power of the same base: 10 = 100 2 10 = 10 (write both sides of the equation as a power of 10) Since eponential functions are one-to-one, we can now assume that since both sides of the equation are epressed as powers of the same base, the eponents must be equal, which allows us to conclude that = 2 But this technique requires us to be able to write both sides of the equation in terms of the same base. In order to learn how to solve eponential equations involving epressions that don t allow us to do this so easily, let s investigate another method of solving eponential equations. Recall that the log-of-powers law allows us to move powers out of the eponent. Thus, let s utilize a logarithmic function to bring the out of the eponent: 10 = 100 log( 10 ) = log( 10 ) 2 (take the common logarithm of both sides of the equation) log(10) = log( 100) (use the log-of-powers law) = 2 (since log(10) = 1 and log(100) = 2) EXAMPLE: Solve the equation 8 = 16 for. Obtain an eact solution. Since both 8 and 16 can be epressed as powers of the same base (2) we can solve the equation using the same base method.
2 8 = 16 ( 2 ) 2 2 = 2 = = (since eponential functions are one-to-one, we can set the eponents equal) But, again, we are lucky to be able to use this method, since both sides of the equation can be epressed as powers of the same base. Another way to solve this equation is to use logarithms. We use the natural logarithm in this case since it is easiest to write (only two letters!) and it is the usually easiest logarithm to estimate on calculators. 8 = 16 ( 8 ) ln ( 16) ln(8) = ln(16) ln(8) ln(16) ln(8) ln(8) = = = = ln(16) ln(8) ln 2 ln 2 l n(2) ln(2) (this value could be easily calculated on a calclator; we will simplify the epression using the log-of-powers law.) In general, when solving eponential equations we are not usually able to use the same base method but we can ALWAYS use the method involving logarithms. The net few eamples we NEED logarithms to solve the equations. We will use the natural logarithm here for the same reasons given in the eample above. Although ANY logarithm can be used, the natural logarithm (or common logarithm) is recommended.
EXAMPLE: Solve the equation 19 + 5 = for. Obtain an eact solution. 19 + 5 = 5 = 1 ( 5 ) ln ( 1) ln(5) = ln(1) = ln(1) ln(5) (first, isolate the eponential epression) Although our solution isn t very user friendly (most of us don t have much intuition about the number ln(1) ), we were asked to find and eact solution, and we can t epress this ln(5) number in a simpler way. So we need to leave this as our solution. If you want to approimate it, that is fine, but the approimation must be given in addition to the eact solution, NOT instead of the eact solution. EXAMPLE: Solve the equation p 5 = 21 for p. Obtain an eact solution. p = p 26 p 26 p 1 ln ( 2 ) p ln 1 ln ( 2 ) ln 1 ( 2 ) p ln 5 21 = p = ln2 ln ln1 ln (once we have the eponential epression isolated, we apply the natural logarithm to both sides)
EXAMPLE: Solve the equation t + 1 8 = 2 for t. Obtain an eact solution. t + 1 8 = 2 t + 1 = t + 1 2 8 = t + 1 ( ) ln (t + 1)ln = ln ln() t + 1= ln() ln() t = 1 ln() t = 1 ln() ln() ( 1) (once we have the eponential epression isolated, we apply the natural logarithm to both sides) EXAMPLE: Solve the equation 25 = 7 for. Obtain an eact solution. 25 = 7 7 2 5 = 5 7 2 5 7 2 5 7 ( ) ln ( 2) ln 5 7 ( ) ln ( 2) 7 ( 2 ) ln 5 ln = = ln ln(7) ln(2) = ln(5) ln()
5 EXAMPLE: Solve the equation = 11 for. Obtain an eact solution. EXAMPLE: Solve the equation 1 = for. Obtain an eact solution. 5 2 EXAMPLE: Solve the equation 2 = 5e for. Obtain an eact solution. EXAMPLE: If y = f = 1, find f ( y) =. 1 EXAMPLE: If a = g( b) = ( b), find b g ( a) log 2 =.
1 EXAMPLE: If p = k( m) = log ( m+ 2), find m k ( p) =. 6 EXAMPLE: If 5 r 1 w= h r =, find r = h ( w). APPLICATIONS INVOVLVING EXPONENTIAL FUNCTIONS EXAMPLE: Suppose that a population grows according to the model t represents years after January 1, 2002. Pt () = Ae where a. What is the doubling-time for the population? b. How long will it take for the population to increase 10%? SOLUTIONS: a. In order to find the doubling-time, we need to determine how long it takes for the population to double. Since the population has initial size A, we need to find t such that Pt () = 2A. (Although we don t know what number A represents, we can still determine when the population is 2A.) Pt () = 2A Ae = 2A e = 2 ( e ) ln ( 2) ( e) t ln ( 2) ln = ln 2 0.07 = ln 2 t = 9.9 0.07
7 Therefore, the doubling-time is about 9.9 years. (One thing worth keeping in mind about the doubling-time is that it represents the time it takes for the population to double no matter when you start counting. So it takes the same 9.9 years for the population to go from 2A to A or from 5A to 10A.) b. If it grows 10%. then the population will be A +.10A = 1.10A, so we need to find the time t such that Pt = 1.10A: Pt () = 1.10A Ae = 1.10A e = 1.10 ( e ) ln( 1.10) ln = ( e) ln = ln 1.10 ln( 1.10) t = 1.6 Therefore, it takes about 1.6 years for the population to grow by 10%. 0.07 Try this one yourself and check your answer. Iodine-11 was one of the radioactive substances released into the atmosphere during the Chernobyl disaster in the former Soviet Union in 1987. If iodine-11 decays at the continuous daily rate of 8.66%, find its half-life. If you have A grams of iodine-11 then the amount left after t days is given by the function It () = Ae 0.0866t To find the half-life, we need to determine how long it takes for the amount of iodine to shrink by 50%, so we need to solve I () t = 0.5A for t. It () = 0.5A 0.0866t Ae = 0.5A 0.0866t e = 0.5 0.0866t ( e ) ln ( 0.5) 0.0866t = ln 0.5 So the half-life of iodine-11 is about 8 days. ln 0.5 t = 8 0.0866
8 EXAMPLE: The half-life of radium-226 is 1620 years. What is the continuous annual rate of decay for radium-226? In order to answer the question, we need to solve the equation below for k. Ae k 1620 1620k = 0.5A e = 0.5 1620k ( e ) ln ( 0.5) ( 5) 1620k = ln 0. ln 0.5 k = 0.00028 1620 So the continuous annual decay rate for radium-226 is about 0.0%. EXAMPLE: Suppose that the population of the island-nation Enlargia is currently growing at the simple annual rate of.%. At what continuous annual rate is the population growing? Since the population of Enlargia is growing at the simple annual rate of.%, the function f( t) = A (1.0) t models the population t years from now, where A is the current population. If we want the continuous annual growth rate, we will need to find k so that kt f () t = Ae. Since we want both f( t) = A (1.0) t and f () t = kt Ae to represent the Enlargia s population, we need them to be equivalent: A (1.0) = Ae t t k t k t (1.0) = e k (Here we've set the growth factors for the two functions equal. 1.0 = e In the future, we can skip the previous steps and start here.) ln(1.0) = k k 0.025 Therefore, Enlargia s.% simple annual growth rate is approimately equivalent to a.25% continuous annual growth rate. Notice that the continuous annual rate is smaller than the simple annual rate since the continuous rate allows interest to earn interest.
9 Try this one yourself and check your answer. The population of the island-nation of Hypothetica is decreasing at the simple annual rate of %. At what continuous annual rate is the population of Hypothetica decreasing? To answer the question we need to fink k so that 0.96 0.96 = k e ln(0.96) = k k 0.008 k = e : Therefore, Hypothetica s % simple annual decay rate is approimately equivalent to a.08% continuous annual decay rate. EXAMPLE: Bonny and Sunny deposit $800 simultaneously at two different banks. a. If Bonny s money earns 5% simple annual interest, find a rule for the function b that gives the amount in Bonny s account t years after making the deposit. b. At Sunny s bank, interest is compounded continuously. If Sunny earns the same annual yield as does Bonny (i.e., at the end of each year, their accounts have the same balance which implies that their accounts always have the same balance), what continuous annual interest rate does Sunny s bank give? a. The desired function is bt = 800(1.05) t. b. Since Sunny also deposits $800, if her continuous interest rate is k then the value of her account is given by the function st () = 800e kt Since Sunny s annual yield is the same as Bonny s, we know that use this fact to find k: st () = bt (). We can
st () = bt () kt t 800e = 800(1.05) kt e t = (1.05) k e = 1.05 k = ln(1.05) 0.88 10 So Sunny s bank gives about.88% continuous annual interest.