2 ln(x+1)ln(4)andg(x) =2 ln(x)ln(2), we see thegraphs intersect only atx= 1 +√3≈2.732. The solutionx= 1−√3<0, which means ifsubstituted into the original equation, the term 2 log2(1−√3)is undefined.y=f(x) = 1 + 2 log4(x+ 1) andy=g(x) = 2 log2(x)
462Exponential and Logarithmic FunctionsIf nothing else, Example6.4.1demonstrates 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 Example6.3.1, much can belearned from checking all of the answers in Example6.4.1analytically. We leave this to the readerand turn our attention to inequalities involving logarithmic functions. Since logarithmic functionsare continuous on their domains, we can use sign diagrams.Example 6.4.2.Solve the following inequalities. Check your answer graphically using a calculator.1.1ln(x) + 1≤12. (log2(x))2<2 log2(x) + 33.xlog(x+ 1)≥xSolution.1. We start solving1ln(x)+1≤1 by getting 0 on one side of the inequality:1ln(x)+1−1≤0.Getting 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. Sincee≈2.718>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= 1.0(+)1e‽(−)10(+)y=f(x) =1ln(x)+1andy=g(x) = 12Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation.
6.4 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. 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