Solving Logarithmic Equations, precalculus lesson

If the pH of a solution is 4.5, its hydrogen ion concentration $x$ in moles per liter satisfies $-\log x = 4.5$ . From $\log x = -4.5$ the definition of the logarithm gives $x = 10^{-4.5} \approx 3.2 \times 10^{-5}$ . Many logarithmic equations are this quick. The catch is that the usual steps can also produce numbers that do not belong, so every answer needs a check.

Key terms

Exponential form conversion

Definition Let $b > 0$ with $b \ne 1$ , let $c$ be a real number and let $A$ be an expression in $x$ . Then

\begin{gathered} \log_b A = c \\[0.4em] \text{exactly when} \quad A = b^c. \end{gathered}

This is the definition of the logarithm, so the conversion is an equivalent step. Since $b^c > 0$ , every solution of $A = b^c$ automatically makes the argument $A$ positive.

In plain words Once a single logarithm equals a number, rewrite the equation as a power. The base of the logarithm becomes the base of the power, and the number on the other side becomes the exponent.

Example The equation $\log_3(2x + 1) = 4$ becomes $2x + 1 = 3^4 = 81$ , so $x = 40$ . As a check, $\log_3 81 = 4$ . In the same way $\ln(x - 2) = 1.5$ becomes $x - 2 = e^{1.5}$ , so

x = 2 + e^{1.5} \approx 6.4817.

Domain validation

Definition The domain of a logarithmic equation is the set of real numbers $x$ at which every part of the original equation is defined. For each logarithm this means that its argument is positive and its base is positive and not $1$ . Domain validation means finding this set before solving and accepting a candidate only if it lies in the set and satisfies the original equation.

In plain words Write down the allowed values of $x$ first. A number outside them is never a solution, however it was found.

Example In $\log_2(x - 1) + \log_2(x + 2) = 2$ the arguments need $x - 1 > 0$ and $x + 2 > 0$ . Both hold exactly when $x > 1$ , so the domain is $(1, \infty)$ . Bases count too. In $\log_x(12 - x) = 2$ the base needs $x > 0$ and $x \ne 1$ , and the argument needs $x < 12$ . Converting gives $x^2 = 12 - x$ , with roots $3$ and $-4$ , and only $3$ lies in the domain. Indeed $\log_3 9 = 2$ .

Equal logarithms

Definition Let $b > 0$ with $b \ne 1$ . For positive $A$ and $B$ ,

\begin{gathered} \log_b A = \log_b B \\[0.4em] \text{exactly when} \quad A = B, \end{gathered}

because $\log_b$ is one-to-one on $(0, \infty)$ . The equation $A = B$ by itself does not require $A$ and $B$ to be positive, so its solutions are only candidates until the domain is checked.

In plain words When one logarithm equals another with the same base, drop the logarithms and set the arguments equal. Then make sure both arguments are positive at each answer.

Example Solve $\ln\left(x^2 - 3\right) = \ln(2x)$ . Setting the arguments equal gives $x^2 - 2x - 3 = 0$ , so $x = 3$ or $x = -1$ . At $x = -1$ both arguments equal $-2$ . They match, yet neither logarithm exists, so $-1$ is rejected. At $x = 3$ both sides are $\ln 6$ , so the only solution is $x = 3$ .

Condensing first

Definition When an equation contains several logarithms with the same base, the product, quotient and power rules combine them into a single logarithm, which can then be converted to exponential form. For $M > 0$ and $N > 0$ ,

\begin{aligned} \log_b M + \log_b N &= \log_b(MN), \\[0.4em] \log_b M - \log_b N &= \log_b \frac{M}{N}. \end{aligned}

The combined logarithm $\log_b(MN)$ is defined whenever $MN > 0$ , which also allows $M < 0$ and $N < 0$ , and the same is true of $\log_b \frac{M}{N}$ . So condensing can enlarge the domain.

In plain words Gather the logarithms into one, rewrite it as a power and solve. Then go back to the domain of the original equation, not the condensed one.

Example Return to $\log_2(x - 1) + \log_2(x + 2) = 2$ , with domain $x > 1$ . Condensing and converting give

\begin{gathered} (x - 1)(x + 2) = 2^2 = 4, \\[0.4em] x^2 + x - 6 = 0, \\[0.4em] (x - 2)(x + 3) = 0. \end{gathered}

The candidate $x = 2$ lies in the domain, and $\log_2 1 + \log_2 4 = 0 + 2 = 2$ , so it is a solution. The candidate $x = -3$ lies outside the domain.

Extraneous solutions

Definition An extraneous solution is a number produced by the solving steps that satisfies a rewritten equation but not the original one. With logarithms the usual cause is a step that enlarges the domain, such as condensing $\log_b M + \log_b N$ into $\log_b(MN)$ or rewriting $2 \log_b M$ as $\log_b\left(M^2\right)$ , which is defined for every $M \ne 0$ . Such a step keeps every solution but may add new ones outside the original domain.

In plain words An extraneous candidate fits the simplified equation, so the algebra gives no warning. Only a check in the original equation catches it.

Example At $x = -3$ the condensed equation above is true, since $(-4)(-1) = 4$ , but the original equation needs $\log_2(-4)$ , which is undefined. On a graph, $y = \log_2\big((x - 1)(x + 2)\big)$ has two branches, one for $x < -2$ and one for $x > 1$ , and both meet the line $y = 2$ . The original left side has only the right branch, so it meets the line once. The power rule can cause the same thing. In $2 \log x = \log(3x + 4)$ rewriting the left side as $\log\left(x^2\right)$ leads to $x^2 - 3x - 4 = 0$ , so $x = 4$ or $x = -1$ . The original equation needs $x > 0$ , so $-1$ is extraneous, and $x = 4$ checks, since $2 \log 4 = \log 16$ .

Steps that lose solutions

Definition A step that shrinks the domain can lose solutions, and a lost solution never comes back in the check. The main example is expanding with the power rule, replacing $\log_b\left(x^2\right)$ , which is defined for every $x \ne 0$ , by $2 \log_b x$ , which is defined only for $x > 0$ . The safe form is

\log_b\left(x^2\right) = 2 \log_b \lvert x \rvert, \qquad x \ne 0.

In plain words Extra candidates can be checked and thrown away, but missing ones cannot be checked at all. Keep the original form, or use absolute values, when a step would cut the domain down.

Example The equation $\ln\left(x^2\right) = 2$ converts to $x^2 = e^2$ , so $x = e$ or $x = -e$ . Both work, since $(-e)^2 = e^2$ . Rewriting the equation first as $2 \ln x = 2$ gives only $x = e$ and loses the solution $-e \approx -2.718$ .

Different bases in one equation

Definition The product, quotient and power rules combine logarithms only when they share a base. If the bases differ, rewrite the logarithms in one base with the change of base formula

\log_b x = \frac{\log_a x}{\log_a b},

valid for positive bases $a$ and $b$ other than $1$ and for $x > 0$ .

In plain words Bring every logarithm to one base, often the smallest base in the equation, and then solve as before.

Example Solve $\log_4 x = \log_2(x - 2)$ , whose domain is $x > 2$ . Change of base gives

\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2} \log_2 x,

so the equation becomes

\begin{aligned} \log_2 x &= 2 \log_2(x - 2) \\[0.4em] &= \log_2\big((x - 2)^2\big) \end{aligned}

and then $x = (x - 2)^2$ . That is $x^2 - 5x + 4 = 0$ , with roots $1$ and $4$ . The root $1$ lies outside the domain. It slipped in when $2 \log_2(x - 2)$ became $\log_2\big((x - 2)^2\big)$ , a step that is correct for $x > 2$ but also allows $x < 2$ . The root $4$ checks, since $\log_4 4 = 1 = \log_2 2$ .

Check yourself

Solve $\log_5(3x - 1) = 2$ .

Show the answer

Converting gives $3x - 1 = 5^2 = 25$ , so $x = \frac{26}{3}$ . At this value the argument is $25$ , which is positive, and $\log_5 25 = 2$ .

Solve $\log x + \log(x - 21) = 2$ .

Show the answer

The domain is $x > 21$ . Condensing and converting give $x(x - 21) = 100$ , so $x^2 - 21x - 100 = 0$ and $(x - 25)(x + 4) = 0$ . The candidate $-4$ lies outside the domain. The solution is $x = 25$ , and indeed $\log 25 + \log 4 = \log 100 = 2$ .

Solve $\log\left(x^2\right) = 4$ .

Show the answer

Converting gives $x^2 = 10^4$ , so $x = 100$ or $x = -100$ . Both are in the domain $x \ne 0$ , and both satisfy the equation. Rewriting the left side as $2 \log x$ first would lose the solution $-100$ .

Solving Logarithmic Equations

Key terms

Exponential form conversion

Domain validation

Equal logarithms

Condensing first

Extraneous solutions

Steps that lose solutions

Different bases in one equation

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus