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Mathematics LibreTexts

6.3: Exponential Equations and Inequalities

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In Section ??? we solved equations and inequalities involving exponential functions using one of two basic strategies. We now turn our attention to equations and inequalities involving logarithmic functions, and not surprisingly, there are two basic strategies to choose from. For example, suppose we wish to solve log2(x)=log2(5). Theorem ??? tells us that the \textit{only} solution to this equation is x=5. Now suppose we wish to solve log2(x)=3. If we want to use Theorem ???, we need to rewrite 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 that x=8. However, we could have arrived at the same answer, in fewer steps, by using Theorem ??? to rewrite the equation log2(x)=3 as 23=x, or x=8. We summarize the two common ways to solve log equations below.

Steps for Solving an Equation involving Logarithmic Functions

  1. Isolate the logarithmic function.
  2. If convenient, express both sides as logs with the same base and equate the arguments of the log functions.
  3. Otherwise, rewrite the log equation as an exponential equation.

Example 6.3.1:

Solve the following equations. Check your solutions graphically using a calculator.

  1. log117(13x)=log117(x23)
  2. 2ln(x3)=1
  3. log6(x+4)+log6(3x)=1
  4. log7(12x)=1log7(3x)
  5. log2(x+3)=log2(6x)+3
  6. 1+2log4(x+1)=2log2(x)

Solution

  1. Since we have the same base on both sides of the equation log117(13x)=log117(x23), we equate what's inside the logs to get 13x=x23. Solving x2+3x4=0 gives x=4 and x=1. To check these answers using the calculator, we make use of the change of base formula and graph f(x)=ln(13x)ln(117) and g(x)=ln(x23)ln(117) and we see they intersect only at x=4. To see what happened to the solution x=1, we substitute it into our original equation to obtain log117(2)=log117(2). While these expressions look identical, neither is a real number,\footnote{They do, however, represent the same \textbf{family} of complex numbers. We stop ourselves at this point and refer the reader to a good course in Complex Variables.} which means x=1 is not in the domain of the original equation, and is not a solution.

\item Our first objective in solving 2ln(x3)=1 is to isolate the logarithm. We get ln(x3)=1, which, as an exponential equation, is e1=x3. We get our solution x=e+3. On the calculator, we see the graph of f(x)=2ln(x3) intersects the graph of g(x)=1 at x=e+35.718.

  1. \item We can start solving log6(x+4)+log6(3x)=1 by using the Product Rule for logarithms to rewrite the equation as log6[(x+4)(3x)]=1. Rewriting this as an exponential equation, we get 61=(x+4)(3x). This reduces to x2+x6=0, which gives x=3 and x=2. Graphing y=f(x)=ln(x+4)ln(6)+ln(3x)ln(6) and y=g(x)=1, we see they intersect twice, at x=3 and x=2.
  1. \item Taking a cue from the previous problem, we begin solving log7(12x)=1log7(3x) by first collecting the logarithms on the same side, log7(12x)+log7(3x)=1, and then using the Product Rule to get log7[(12x)(3x)]=1. Rewriting this as an exponential equation gives 71=(12x)(3x) which gives the quadratic equation 2x27x4=0. Solving, we find x=12 and x=4. Graphing, we find y=f(x)=ln(12x)ln(7) and y=g(x)=1ln(3x)ln(7) intersect only at x=12. Checking x=4 in the original equation produces log7(7)=1log7(1), which is a clear domain violation.
  2. \item Starting with log2(x+3)=log2(6x)+3, we gather the logarithms to one side and get log2(x+3)log2(6x)=3. We then use the Quotient Rule and convert to an exponential equation log2(x+36x)=323=x+36x This reduces to the linear equation 8(6x)=x+3, which gives us x=5. When we graph f(x)=ln(x+3)ln(2) and g(x)=ln(6x)ln(2)+3, we find they intersect at x=5.
  1. \item Starting with 1+2log4(x+1)=2log2(x), we gather the logs to one side to get the equation 1=2log2(x)2log4(x+1). Before we can combine the logarithms, however, we need a common base. Since 4 is a power of 2, we use change of base to convert log4(x+1)=log2(x+1)log2(4)=12log2(x+1) Hence, our original equation becomes

1=2log2(x)2(12log2(x+1))1=2log2(x)log2(x+1)1=log2(x2)log2(x+1)Power Rule1=log2(x2x+1)Quotient Rule

Rewriting this in exponential form, we get x2x+1=2 or x22x2=0. Using the quadratic formula, we get x=1±3. Graphing f(x)=1+2ln(x+1)ln(4) and g(x)=2ln(x)ln(2), we see the graphs intersect only at x=1+32.732. The solution x=13<0, which means if substituted into the original equation, the term 2log2(13) is undefined.

If nothing else, Example ??? demonstrates the importance of checking for extraneous solutions\footnote{Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation.} when solving equations involving logarithms. Even though we checked our answers graphically, extraneous solutions are easy to spot - any supposed solution which causes a negative number inside a logarithm needs to be discarded. As with the equations in Example ???, much can be learned from checking all of the answers in Example ??? analytically. We leave this to the reader and turn our attention to inequalities involving logarithmic functions. Since logarithmic functions are continuous on their domains, we can use sign diagrams.

Example 6.3.2:

Solve the following inequalities. Check your answer graphically using a calculator.

  1. 1ln(x)+11
  2. (log2(x))2<2log2(x)+3
  3. xlog(x+1)x

Solution

\item We start solving 1ln(x)+11 by getting 0 on one side of the inequality: 1ln(x)+110. Getting a common denominator yields 1ln(x)+1ln(x)+1ln(x)+10 which reduces to ln(x)ln(x)+10, or ln(x)ln(x)+10. We define r(x)=ln(x)ln(x)+1 and set about finding the domain and the zeros of r. Due to the appearance of the term ln(x), we require x>0. In order to keep the denominator away from zero, we solve ln(x)+1=0 so ln(x)=1, so x=e1=1e. Hence, the domain of r is (0,1e)(1e,). To find the zeros of r, we set r(x)=ln(x)ln(x)+1=0 so that ln(x)=0, and we find x=e0=1. In order to determine test values for r without resorting to the calculator, we need to find numbers between 0, 1e, and 1 which have a base of e. Since e2.718>1, 0<1e2<1e<1e<1<e. To determine the sign of r(1e2), we use the fact that ln(1e2)=ln(e2)=2, and find r(1e2)=22+1=2, which is (+). The rest of the test values are determined similarly. From our sign diagram, we find the solution to be (0,1e)[1,). Graphing f(x)=1ln(x)+1 and g(x)=1, we see the the graph of f is below the graph of g on the solution intervals, and that the graphs intersect at x=1.

  1. \item Moving all of the nonzero terms of (log2(x))2<2log2(x)+3 to one side of the inequality, we have (log2(x))22log2(x)3<0. Defining r(x)=(log2(x))22log2(x)3, we get the domain of r is (0,), due to the presence of the logarithm. To find the zeros of r, we set r(x)=(log2(x))22log2(x)3=0 which results in a `quadratic in disguise.' We set u=log2(x) so our equation becomes u22u3=0 which gives us u=1 and u=3. Since u=log2(x), we get log2(x)=1, which gives us x=21=12, and log2(x)=3, which yields x=23=8. We use test values which are powers of 2: 0<14<12<1<8<16, and from our sign diagram, we see r(x)<0 on (12,8). Geometrically, we see the graph of f(x)=(ln(x)ln(2))2 is below the graph of y=g(x)=2ln(x)ln(2)+3 on the solution interval.
  2. We begin to solve xlog(x+1)x by subtracting x from both sides to get xlog(x+1)x0. We define r(x)=xlog(x+1)x\(andduetothepresenceofthelogarithm,werequire\(x+1>0, or x>1. To find the zeros of r, we set r(x)=xlog(x+1)x=0. Factoring, we get x(log(x+1)1)=0, which gives x=0 or log(x+1)1=0. The latter gives log(x+1)=1, or x+1=101, which admits x=9. We select test values x so that x+1 is a power of 10, and we obtain 1<0.9<0<101<9<99. Our sign diagram gives the solution to be (1,0][9,). The calculator indicates the graph of y=f(x)=xlog(x+1) is above y=g(x)=x on the solution intervals, and the graphs intersect at x=0 and x=9.

Our next example revisits the concept of pH as first introduced in the exercises in Section ???.

Example 6.3.1: Fish Talk pH

In order to successfully breed Ippizuti fish the pH of a freshwater tank must be at least 7.8 but can be no more than 8.5. Determine the corresponding range of hydrogen ion concentration, and check your answer using a calculator.

Solution

Recall from Exercise ??? in Section ??? that pH=log[H+] where [H+] is the hydrogen ion concentration in moles per liter. We require 7.8log[H+]8.5 or 7.8log[H+]8.5. To solve this compound inequality we solve 7.8log[H+] and log[H+]8.5 and take the intersection of the solution sets.\footnote{Refer to page \pageref{intersectionunion} for a discussion of what this means.} The former inequality yields 0<[H+]107.8 and the latter yields [H+]108.5. Taking the intersection gives us our final answer 108.5[H+]107.8. (Your Chemistry professor may want the answer written as 3.16×109[H+]1.58×108.) After carefully adjusting the viewing window on the graphing calculator we see that the graph of f(x)=log(x) lies between the lines y=7.8 and y=8.5 on the interval [3.16×109,1.58×108].

We close this section by finding an inverse of a one-to-one function which involves logarithms.

Example 6.3.1:

The function f(x)=log(x)1log(x) is one-to-one. Find a formula for f1(x) and check your answer graphically using your calculator.

Solution

We first write y=f(x) then interchange the x and y and solve for y.

y=f(x)y=log(x)1log(x)x=log(y)1log(y)Interchange x and y.x(1log(y))=log(y)xxlog(y)=log(y)x=xlog(y)+log(y)x=(x+1)log(y)xx+1=log(y)y=10xx+1Rewrite as an exponential equation.

We have f1(x)=10xx+1. Graphing f and f1 on the same viewing window yields


6.3: Exponential Equations and Inequalities is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by LibreTexts.

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