Logarithmic Equations Solving Logarithmic Equations Logarithmic Equations Solving Logarithmic Equations Logarithmic Functions This module explores more detailed solutions of exponential and logarithmic equations. Logarithmic Equations 8/15/2013

Solving Logarithmic Equations Solving Equations Solving Logarithmic Equations Solve 1. log3 x = –3 With inverse function: With definition: log3 x = –3 log3 x = –3 Exponentiate: Apply Definition: 3 log 3 x = 3 –3 3 –3 = x Solving Equations Solving a simple logarithm equation is done very much like solving an exponential equation: apply the inverse function to reduce the equation to a linear or polynomial equation. In the first example we simply exponentiate both sides of the equation, which gets rid of the logarithm, leaving a simple linear equation to solve. The solution is then trivial. The alternative solution simply applies the definition of logarithm directly, giving a simple linear equation, with the solution already isolated. In the second example we again exponentiate both sides of the equation to produce a radical equation. Solving by raising both sides to the same power produces a simple exponential equation that is solved by applying the 1-1 property of exponential functions. 27 1 = x 27 1 = x 27 1 { } Solution set: 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Logarithmic Equations Solving Equations Solving Logarithmic Equations Solve 2. = x 6  4 log36 This is already solved for x , so simplify 6  4 36 x = By definition the logarithm (i.e. x) is the power of the base that yields 6  4 61/4 62x = 4 1 = 2x WHY ? Solving Equations Solving a simple logarithm equation is done very much like solving an exponential equation: apply the inverse function to reduce the equation to a linear or polynomial equation. In the first example we simply exponentiate both sides of the equation, which gets rid of the logarithm, leaving a simple linear equation to solve. The solution is then trivial. The alternative solution simply applies the definition of logarithm directly, giving a simple linear equation, with the solution already isolated. In the second example we again exponentiate both sides of the equation to produce a radical equation. Solving by raising both sides to the same power produces a simple exponential equation that is solved by applying the 1-1 property of exponential functions. 8 1 = x { } 8 1 Solution set: 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Logarithmic Equations Solving Equations Solving Logarithmic Equations Solve 3. log x 5 = 4 By inverse function: By definition: Exponentiate: Apply Definition: x log x 5 = x4 loga x = b iffi ab = x 5 = x4 x4 = 5 4  x4 5 = x  5 4 = Solving Equations This example starts by exponentiating both sides, giving a trivial polynomial equation, which is solved by finding the fourth root of both sides. The alternate solution applies the 1-1 property of logarithm functions, to produce the same power function. Note that the solution can’t be negative, since that would force the logarithm base to be negative – which would violate the definition of logarithm ! Hence only the positive root can be considered. =  x  x , for x > 0 = –x , for x < 0 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Logarithmic Equations Solving Equations Solving Logarithmic Equations Solve 3. log x 5 = 4 4  5 –x , for x < 0 x , for x > 0 = x  5 4 = OR { }  5 4 Solution set: Question: Since x2 = 5 yields x =   5 2 then Solving Equations This example started by exponentiating both sides, giving a trivial polynomial equation, which is solved by finding the fourth root of both sides. The alternate solution applies the 1-1 property of logarithm functions, to produce the same power function. Note that the solution can’t be negative, since that would force the logarithm base to be negative – which would violate the definition of logarithm ! Hence only the positive root can be considered. Note that x2 = 5 yields two real roots, only one of which makes sense. This differs from x4 = 5 which yields four roots, two real and two imaginary. Since we are dealing with real solutions of real-valued functions, we cannot admit x4 = 5 without some restrictions on roots. why doesn’t x4 = 5 yield x =   5 4 ? Note: then log x 5 is defined x  5 – = If with a negative base !! 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Logarithmic Equations Solving Equations Solving Logarithmic Equations Solve 4. log 3 (x2 + 5) = 2 By inverse function: By definition: Exponentiate: ( 2 is the power of 3 that yields x2 + 5 ) 3 log3 (x2 + 5) = 3 2 32 = x2 + 5 = x2 + 5 9 9 = x2 + 5 x2 = 4 Solving Equations In the first example we apply the inverse function and exponentiate both sides, producing a polynomial equation. The standard procedure is then followed to solve for x. The alternative solution simply applies the definition of logarithm directly. Clearly the solutions are the same. In the second example we again apply the inverse function, this time by finding the logarithm of both sides. However, we arrive at the embarrassing position of having to find the power of 4 that yields 12, i.e. log412. There is no closed-form solution for this equation. The standard approach to solving this equation is to identify the two sides of the equation as two separate functions: y1 = log412 and y2 = x and then approximate the intersection of the two graphs. OR, we try a change-of-base. With modern calculators, such as the TI-83, TI-84, TI-85 and later models, choose the Math button and then the Solver option. Complete the equation 0 = ...... with the terms of the given equation regrouped on the right. Press the ENTER key to see the result: x = 1.79248125 ... approximately. The calculator has of course solved the equation as described above with iterated approximations. In the next section we will see how to change the base of the logarithm to something else, say log10 or common log. Then, the calculation is very simple: 4 = x2  4 ± = x  4 = x2 = x 2 ± 2 = ± x Solution set: { ± 2 } 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Logarithmic Equations Solving Equations Solving Logarithmic Equations Solve 5. 12 = 4x To remove a variable from an exponent, find the logarithm of the exponential form log 4 12 = log 4 (4x) = x Solving Equations In the first example we apply the inverse function and exponentiate both sides, producing a polynomial equation. The standard procedure is then followed to solve for x. The alternative solution simply applies the definition of logarithm directly. Clearly the solutions are the same. In the second example we again apply the inverse function, this time by finding the logarithm of both sides. However, we arrive at the embarrassing position of having to find the power of 4 that yields 12, i.e. log412. There is no closed-form solution for this equation. The standard approach to solving this equation is to identify the two sides of the equation as two separate functions: y1 = log412 and y2 = x and then approximate the intersection of the two graphs. OR, we try a change-of-base. With modern calculators, such as the TI-83, TI-84, TI-85 and later models, choose the Math button and then the Solver option. Complete the equation 0 = ...... with the terms of the given equation regrouped on the right. Press the ENTER key to see the result: x = 1.79248125 ... approximately. The calculator has of course solved the equation as described above with iterated approximations. In the next section we will see how to change the base of the logarithm to something else, say log10 or common log. Then, the calculation is very simple: Question: Now, how do we find log 4 12 ? 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Logarithmic Equations Solving Equations Solving Logarithmic Equations Remember: ax and logax are inverses 1. To remove a variable from an exponent, find the logarithm of the exponential form 2. To remove a variable from a logarithm, exponentiate Solving Equations These are just review reminders about handling exponential and logarithmic equations. The simple approach is to apply the inverse function in each of these cases, that is, apply the logarithm in the first case, and exponentiate in the second case. 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

One-to-One Property Review Solving Logarithmic Equations Since ax and log a x are 1-1 then 1. If x = y then a x = a y 2. If a x = a y then x = y 3. If x = y then log a x = log a y 4. If log a x = loga y then x = y These facts can be used to solve equations One-to-One Property Review The salient feature to observe about 1-1 functions is that f(x) = f(y) if and only if x = y. The illustration shows what this means in terms of the exponential and logarithm functions. We can use the fact that both the exponential and the logarithm functions are 1-1 to solve exponential and logarithm equations. In the example we can move the variable from the exponent to a multiplier by using the logarithm to act on each side of the equation. This helps us form a linear equation in which we isolate x. By computing log 4 and log 10, in the example, we can solve for x. Note that log 10 = 1, since log without an explicit base is by convention log10. Solving for x gives us x = 0.6021. 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Equations: Examples Solving Logarithmic Equations x 2 7 = 4x – 3 6. Solve Since the exponential function is 1-1 then x2 = 4x – 3 x2 – 4x + 3 = 0 (x – 1)(x – 3) = 0 Apply zero product rule x – 1 = 0 OR x – 3 = 0 Solving Equations In Example 4 we apply the 1–1 property of power functions to set the exponents equal to each other. This produces a simple quadratic equation which is easily solved by factoring and applying the zero product rule. From this we gain two solutions, 1 and 3. We could have applied the inverse function, log7 to isolate the x terms. It turns out we could also have used logarithms of other bases, for example common log and natural log. We would then have to also apply the power rule for logarithms to extract the exponents as multipliers, dividing out the log expression in common with both sides. For example, Applying the power rule, Dividing by log 7, x2 = 4x – 3 , which yields the same solutions. x = 1 OR x = 3 { 1, 3 } Solution set: 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Equations: Examples Solving Logarithmic Equations x 2 7 = 4x – 3 6. Solve { 1, 3 } Solution set: NOTE: Could have applied inverse function x 2 7 log 7 = 7 4x–3 log7 x2 = 4x – 3 x2 – 4x + 3 = 0 (x – 3)(x – 1) = 0 x = 1, 3 Solving Equations In Example 4 we apply the 1–1 property of power functions to set the exponents equal to each other. This produces a simple quadratic equation which is easily solved by factoring and applying the zero product rule. From this we gain two solutions, 1 and 3. We could have applied the inverse function, log7 to isolate the x terms. It turns out we could also have used logarithms of other bases, for example common log and natural log. We would then have to also apply the power rule for logarithms to extract the exponents as multipliers, dividing out the log expression in common with both sides. For example, Applying the power rule, Dividing by log 7, x2 = 4x – 3 , which yields the same solutions. Question: Could we use log or ln instead of log7 ? 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Equations: Examples Solving Logarithmic Equations 7. Solve 5 ln x = 10 By definition 2 is the power of the base that yields x ln x = 2 Thus x = e2 ≈ (2.71828)2 ≈ 7.38906 Solving Equations In Example 5 we isolate ln x and then use the definition of logarithm to isolate x as the square of e. This we evaluate as approximately 7.38906. This value is only approximate, since e is irrational (in fact, transcendental) so that algebraic expressions in decimal fractions must be approximate. For Example 6 we employ the definition of logarithm to extract the expression in x from the logarithm. This linear equation is trivial to solve. Solution set: { ~ 7.38906 } 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Equations: Examples Solving Logarithmic Equations 8. Solve log 3 (1 – x) = 1 By definition: 31 = 1 – x Thus x = 1 – 3 = –2 Solution set: { –2 } Solving Equations In Example 5 we isolate ln x and then use the definition of logarithm to isolate x as the square of e. This we evaluate as approximately 7.38906. This value is only approximate, since e is irrational (in fact, transcendental) so that algebraic expressions in decimal fractions must be approximate. For Example 6 we employ the definition of logarithm to extract the expression in x from the logarithm. This linear equation is trivial to solve. 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Equations: Examples Solving Logarithmic Equations 9. Solve ln (x2 – 4) – ln (x + 2) = ln (3 – x) Using the quotient rule ln x2 – 4 x + 2 ( ) = ln (3 – x) Since the logarithm function is 1-1, then x2 – 4 x + 2 = 3 – x Solving Equations Example 7 requires the use of the quotient rule for logarithms, followed by the application of the 1–1 property of logarithms. The rational expression so produced can be cleared of fractions and the resulting quadratic equations can be solved by factoring or by used of the quadratic formula. Factoring is much simpler and faster and so is used to set up the application of the zero product property. The two solutions are then –2 and 5/2. However, for x = –2 we note that ln(x2 – 4) and ln(x + 2) do not exist, since the natural logarithm is not defined at 0. So –2 is not a solution, leaving 5/2 as the only solution. This is easily seen by noting that the domain of any logarithm function is the set of all positive real numbers and the range is the set of all real numbers. x2 – 4 = (x + 2)(3 – x) = 3x – x2 + 6 – 2x 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Equations: Examples Solving Logarithmic Equations 9. Solve ln (x2 – 4) – ln (x + 2) = ln (3 – x) x2 – 4 = 3x – x2 + 6 – 2x 2x2 – x – 10 = 0 = (2x – 5)(x + 2) 2x – 5 = 0 OR x + 2 = 0 x = 5/2 OR x = –2 Solving Equations Example 7 requires the use of the quotient rule for logarithms, followed by the application of the 1–1 property of logarithms. The rational expression so produced can be cleared of fractions and the resulting quadratic equations can be solved by factoring or by used of the quadratic formula. Factoring is much simpler and faster and so is used to set up the application of the zero product property. The two solutions are then –2 and 5/2. However, for x = –2 we note that ln(x2 – 4) and ln(x + 2) do not exist, since the natural logarithm is not defined at 0. So –2 is not a solution, leaving 5/2 as the only solution. This is easily seen by noting that the domain of any logarithm function is the set of all positive real numbers and the range is the set of all real numbers. Solution set: { 5/2 } Why not –2 ? Question: What are the domain and range of the ln function ? 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Logarithmic Equations Solving Equations Solving Logarithmic Equations Quotient Example 10. Solve for x : log (x + 3) log (x + 1) = 2 log (x + 3) log (x + 1) = 2 log (x + 1) = 2 WHY? x + 3 = (x + 1) 2 WHY? = x2 + 2x+ 1 Properties of Logarithms: The Quotient Rule Examples In Example 1, we use a simple quotient, replacing the x and y in statement of the rule with 9 and 4, respectively. In Example 2, we again apply the quotient rule to produce a difference of logarithms. In this case, the second logarithm is reducible to a constant. In Example 3, we solve a logarithmic equation using first the quotient rule and then a special case of the product rule. We then invoke the 1-1 property of the logarithm function to claim the equality of x +3 and (x + 1)2. Expanding the right side and gathering like terms, we form a polynomial equation, which is easily solvable by factoring and applying the zero-product rule. = x2 + x – 2 = (x + 2)(x – 1) Solution set: { –2 , 1 } 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013

Solving Logarithmic Equations Think about it ! 8/15/2013 Logarithmic Equations Logarithmic Equations 8/15/2013