6.4 Logarithmic Equations and Inequalities

1 Logarithmic Equations and Logarithmic Equations and InequalitiesIn Section we solved Equations and Inequalities involving exponential functions using one oftwo basic strategies. We now turn our attention to Equations and Inequalities involving logarithmicfunctions, and not surprisingly, there are two basic strategies to choose from. For example, supposewe wish to solve log2(x) = log2(5). Theorem tells us that theonlysolution to this equationisx= 5. Now suppose we wish to solve log2(x) = 3. If we want to use Theorem , we need torewrite 3 as a logarithm base 2. We can use Theorem to do just that: 3 = log2(23)= log2(8).Our equation then becomes log2(x) = log2(8) so thatx= 8. However, we could have arrived at thesame answer, in fewer steps, by using Theorem to rewrite the equation log2(x) = 3 as 23=x,orx= 8.

2 We summarize the two common ways to solve log Equations for Solving an Equation involving Logarithmic Functions1. Isolate the Logarithmic (a) If convenient, express both sides as logs with the same base and equate the argumentsof the log functions.(b) Otherwise, rewrite the log equation as an exponential the following Equations . Check your solutions graphically using a log117(1 3x) = log117(x2 3)2. 2 ln(x 3) = 13. log6(x+ 4) + log6(3 x) = 14. log7(1 2x) = 1 log7(3 x)5. log2(x+ 3) = log2(6 x) + 36. 1 + 2 log4(x+ 1) = 2 log2(x) Since we have the same base on both sides of the equation log117(1 3x) = log117(x2 3),we equate what s inside the logs to get 1 3x=x2 3. Solvingx2+ 3x 4 = 0 givesx= 4 andx= 1. To check these answers using the calculator, we make use of the changeof base formula and graphf(x) =ln(1 3x)ln(117)andg(x) =ln(x2 3)ln(117)and we see they intersect onlyatx= 4.

3 To see what happened to the solutionx= 1, we substitute it into our originalequation to obtain log117( 2) = log117( 2). While these expressions look identical, neitheris a real number,1which meansx= 1 is not in the domain of the original equation, and isnot a Our first objective in solving 2 ln(x 3) = 1 is to isolate the logarithm. We get ln(x 3) = 1,which, as an exponential equation, ise1=x 3. We get our solutionx=e+ 3. On thecalculator, we see the graph off(x) = 2 ln(x 3) intersects the graph ofg(x) = 1 atx=e+ 3 do, however, represent the samefamilyof complex numbers. We stop ourselves at this point and refer thereader to a good course in Complex and Logarithmic Functionsy=f(x) = log117(1 3x) andy=f(x) = 2 ln(x 3) andy=g(x) = log117(x2 3)y=g(x) = 13. We can start solving log6(x+ 4) + log6(3 x) = 1 by using the Product Rule for logarithms torewrite the equation as log6[(x+ 4)(3 x)] = 1.

4 Rewriting this as an exponential equation,we get 61= (x+ 4)(3 x). This reduces tox2+x 6 = 0, which givesx= 3 andx= (x) =ln(x+4)ln(6)+ln(3 x)ln(6)andy=g(x) = 1, we see they intersect twice, atx= 3 andx= (x) = log6(x+ 4) + log6(3 x) andy=g(x) = 14. Taking a cue from the previous problem, we begin solving log7(1 2x) = 1 log7(3 x) byfirst collecting the logarithms on the same side, log7(1 2x) + log7(3 x) = 1, and then usingthe Product Rule to get log7[(1 2x)(3 x)] = 1. Rewriting this as an exponential equationgives 71= (1 2x)(3 x) which gives the quadratic equation 2x2 7x 4 = 0. Solving, we findx= 12andx= 4. Graphing, we findy=f(x) =ln(1 2x)ln(7)andy=g(x) = 1 ln(3 x)ln(7)intersectonly atx= 12. Checkingx= 4 in the original equation produces log7( 7) = 1 log7( 1),which is a clear domain Starting with log2(x+ 3) = log2(6 x) + 3, we gather the logarithms to one side and getlog2(x+ 3) log2(6 x) = 3.

5 We then use the Quotient Rule and convert to an exponentialequationlog2(x+ 36 x)= 3 23=x+ 36 xThis reduces to the linear equation 8(6 x) =x+ 3, which gives usx= 5. When we graphf(x) =ln(x+3)ln(2)andg(x) =ln(6 x)ln(2)+ 3, we find they intersect atx= Logarithmic Equations and Inequalities461y=f(x) = log7(1 2x) andy=f(x) = log2(x+ 3) andy=g(x) = 1 log7(3 x)y=g(x) = log2(6 x) + 36. Starting with 1 + 2 log4(x+ 1) = 2 log2(x), we gather the logs to one side to get the equation1 = 2 log2(x) 2 log4(x+ 1). Before we can combine the logarithms, however, we need acommon base. Since 4 is a power of 2, we use change of base to convertlog4(x+ 1) =log2(x+ 1)log2(4)=12log2(x+ 1)Hence, our original equation becomes1 = 2 log2(x) 2(12log2(x+ 1))1 = 2 log2(x) log2(x+ 1)1 = log2(x2) log2(x+ 1)Power Rule1 = log2(x2x+ 1)Quotient RuleRewriting this in exponential form, we getx2x+1= 2 orx2 2x 2 = 0.

6 Using the quadraticformula, we getx= 1 3. Graphingf(x) = 1 +2 ln(x+1)ln(4)andg(x) =2 ln(x)ln(2), we see thegraphs intersect only atx= 1 + 3 The solutionx= 1 3<0, which means ifsubstituted into the original equation, the term 2 log2(1 3)is (x) = 1 + 2 log4(x+ 1) andy=g(x) = 2 log2(x)462 Exponential and Logarithmic FunctionsIf nothing else, Example demonstrates the importance of checking for extraneous solutions2when solving Equations involving logarithms. Even though we checked our answers graphically,extraneous solutions are easy to spot - any supposed solution which causes a negative numberinside a logarithm needs to be discarded. As with the Equations in Example , much can belearned from checking all of the answers in Example analytically. We leave this to the readerand turn our attention to Inequalities involving Logarithmic functions.

7 Since Logarithmic functionsare continuous on their domains, we can use sign the following Inequalities . Check your answer graphically using a (x) + 1 12. (log2(x))2<2 log2(x) + (x+ 1) We start solving1ln(x)+1 1 by getting 0 on one side of the inequality:1ln(x)+1 1 a common denominator yields1ln(x)+1 ln(x)+1ln(x)+1 0 which reduces to ln(x)ln(x)+1 0,orln(x)ln(x)+1 0. We definer(x) =ln(x)ln(x)+1and set about finding the domain and the zerosofr. Due to the appearance of the term ln(x), we requirex >0. In order to keep thedenominator away from zero, we solve ln(x) + 1 = 0 so ln(x) = 1, sox=e 1=1e. Hence,the domain ofris(0,1e) (1e, ). To find the zeros ofr, we setr(x) =ln(x)ln(x)+1= 0 so thatln(x) = 0, and we findx=e0= 1. In order to determine test values forrwithout resortingto the calculator, we need to find numbers between 0,1e, and 1 which have a base ofe.

8 Sincee >1, 0<1e2<1e<1 e<1< e. To determine the sign ofr(1e2), we use the fact thatln(1e2)= ln(e 2)= 2, and findr(1e2)= 2 2+1= 2, which is (+). The rest of the test valuesare determined similarly. From our sign diagram, we find the solution to be(0,1e) [1, ).Graphingf(x) =1ln(x)+1andg(x) = 1, we see the the graph offis below the graph ofgonthe solution intervals, and that the graphs intersect atx= (+)1e ( )10(+)y=f(x) =1ln(x)+1andy=g(x) = 12 Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original Logarithmic Equations and Inequalities4632. Moving all of the nonzero terms of (log2(x))2<2 log2(x) + 3 to one side of the inequality,we have (log2(x))2 2 log2(x) 3<0. Definingr(x) = (log2(x))2 2 log2(x) 3, we getthe domain ofris (0, ), due to the presence of the logarithm.]

9 To find the zeros ofr, wesetr(x) = (log2(x))2 2 log2(x) 3 = 0 which results in a quadratic in disguise. We setu= log2(x) so our equation becomesu2 2u 3 = 0 which gives usu= 1 andu= 3. Sinceu= log2(x), we get log2(x) = 1, which gives usx= 2 1=12, and log2(x) = 3, which yieldsx= 23= 8. We use test values which are powers of 2: 0<14<12<1<8<16, and from oursign diagram, we seer(x)<0 on(12,8). Geometrically, we see the graph off(x) =(ln(x)ln(2))2is below the graph ofy=g(x) =2 ln(x)ln(2)+ 3 on the solution (+)120( )80(+)y=f(x) = (log2(x))2andy=g(x) = 2 log2(x) + 33. We begin to solvexlog(x+ 1) xby subtractingxfrom both sides to getxlog(x+ 1) x definer(x) =xlog(x+1) xand due to the presence of the logarithm, we requirex+1>0,orx > 1. To find the zeros ofr, we setr(x) =xlog(x+ 1) x= 0. Factoring, we getx(log(x+ 1) 1) = 0, which givesx= 0 or log(x+ 1) 1 = 0.

10 The latter gives log(x+ 1) = 1,orx+ 1 = 101, which admitsx= 9. We select test valuesxso thatx+ 1 is a power of 10,and we obtain 1< <0< 10 1<9<99. Our sign diagram gives the solution tobe ( 1,0] [9, ). The calculator indicates the graph ofy=f(x) =xlog(x+ 1) is abovey=g(x) =xon the solution intervals, and the graphs intersect atx= 0 andx= 9. 1(+)00( )90(+)y=f(x) =xlog(x+ 1) andy=g(x) =x464 Exponential and Logarithmic FunctionsOur next example revisits the concept of pH as first introduced in the exercises in Section order to successfully breed Ippizuti fish the pH of a freshwater tank must beat least but can be no more than Determine the corresponding range of hydrogen ionconcentration, and check your answer using a from Exercise 77 in Section that pH = log[H+] where [H+] is the hydrogenion concentration in moles per liter.