Solving Exponential and Logarithmic Equations, foundations for calculus lesson

A savings balance that grows by a fixed percentage each year raises a natural question. When will it double? The unknown sits in the exponent, so ordinary algebra cannot reach it. The cure is to look at where the unknown is and apply the operation that undoes what is wrapped around it, exponentials undoing logarithms and logarithms undoing exponentials. Some of those steps can produce numbers that do not solve the original equation, so the last step is always a check.

Key terms

Exponential equation

Definition An exponential equation is an equation in which the unknown appears in an exponent, as in $a^{g(x)} = c$ with a base $a > 0$ , $a \ne 1$ . Because $a^{g(x)} > 0$ for every input, the equation $a^{g(x)} = c$ has no solution when $c \le 0$ .

In plain words The unknown is up in the exponent. First isolate the exponential part, then bring the unknown down.

Example To solve $3 \cdot 2^x = 48$ , divide by $3$ to isolate the power, which gives $2^x = 16 = 2^4$ , so $x = 4$ . In contrast, the equation $e^{2x} = -5$ has no solution at all, since $e^{2x}$ is positive for every real $x$ .

One-to-one exponential

Definition For a fixed base $a > 0$ with $a \ne 1$ , the function $x \mapsto a^x$ is one-to-one. So for real numbers $u$ and $v$ ,

a^u = a^v \quad \text{exactly when} \quad u = v.

The condition $a \ne 1$ is needed, because $1^u = 1$ for every $u$ .

In plain words Two powers of the same valid base are equal only when the exponents are equal, so you may set the exponents equal.

Example The equation $5^{x^2} = 5^4$ gives $x^2 = 4$ , so $x = 2$ or $x = -2$ . Both work, since each makes the exponent $4$ . With base $1$ the reasoning fails, because $1^2 = 1^5$ even though $2 \ne 5$ .

Common base

Definition The common base method rewrites both sides of an exponential equation as powers of one base $a > 0$ , $a \ne 1$ , and then uses the one-to-one property to equate the exponents. It works when both sides are known powers of the same number.

In plain words If $4$ and $8$ are both powers of $2$ , write everything with base $2$ and compare exponents.

Example Solve $4^{x + 1} = 8^x$ . Since $4 = 2^2$ and $8 = 2^3$ ,

\begin{gathered} 2^{2(x + 1)} = 2^{3x}, \\[0.4em] \text{so} \quad 2x + 2 = 3x \quad \text{and} \quad x = 2. \end{gathered}

Check in the original equation. The left side is $4^3 = 64$ and the right side is $8^2 = 64$ .

Taking logarithms

Definition If $P > 0$ and $Q > 0$ , then for any valid base $b$ ,

\begin{gathered} P = Q \\[0.4em] \text{exactly when} \\[0.4em] \log_b P = \log_b Q, \end{gathered}

because $\log_b$ is one-to-one on $(0, \infty)$ . Taking a real logarithm of both sides is allowed only when both sides are positive. The answer is usually written exactly with logarithms, and a decimal is given afterward and marked with $\approx$ as an approximation.

In plain words When the two sides cannot be written with a common base, apply a logarithm to both sides and use the power law to bring the exponents down.

Example Solve $3^x = 20$ . Both sides are positive, so take natural logarithms. The power law gives $x \ln 3 = \ln 20$ , hence

x = \frac{\ln 20}{\ln 3} \approx 2.727.

The exact answer is the fraction of logarithms. The decimal is only close, and substituting it does not give exactly $20$ . For instance $3^{2.73} \approx 20.07$ . A harder case has the unknown on both sides. To solve $2^{x + 1} = 3^x$ , take logarithms and collect the $x$ terms.

\begin{gathered} (x + 1) \ln 2 = x \ln 3, \\[0.4em] \text{so} \quad x = \frac{\ln 2}{\ln 3 - \ln 2} \approx 1.710. \end{gathered}

Logarithmic equation

Definition A logarithmic equation is an equation in which the unknown appears inside the argument of a logarithm. For a valid base $a$ and any real number $c$ ,

\begin{gathered} \log_a u = c \\[0.4em] \text{exactly when} \\[0.4em] u = a^c, \end{gathered}

and any $u$ found this way is automatically positive.

In plain words The unknown is trapped inside a logarithm. Rewrite the equation in exponential form to free it.

Example Solve $\log_2(x + 3) = 5$ . In exponential form $x + 3 = 2^5 = 32$ , so $x = 29$ . Check that $\log_2 32 = 5$ , which is true.

Argument restriction

Definition The domain of an equation with logarithms is the set of inputs at which every logarithm appearing in the original equation has a positive argument. A solution must lie in this set. Combining logarithms with the product or quotient law can produce an expression defined on a larger set, so the restriction comes from the original terms, not from the combined form.

In plain words Before you combine anything, write down where each logarithm is defined. That list is the only place a solution can live.

Example Solve $\ln x + \ln(x - 3) = \ln 10$ . The original terms need $x > 0$ and $x - 3 > 0$ , so any solution must satisfy $x > 3$ . On that set the product law gives

\begin{gathered} \ln\left(x(x - 3)\right) = \ln 10, \\[0.4em] \text{so} \quad x^2 - 3x - 10 = 0. \end{gathered}

This factors as $(x - 5)(x + 2) = 0$ , with roots $5$ and $-2$ . Only $x = 5$ satisfies $x > 3$ , and indeed $\ln 5 + \ln 2 = \ln 10$ .

Candidate solution

Definition A candidate solution is a number produced by a chain of solving steps in which each new equation follows from the one before, so every solution of the earlier equation also solves the new one. Combining logarithms with the product law and squaring both sides are steps of this kind, but they are not always reversible, so the new equation may have extra solutions. Under these conditions every true solution appears among the candidates, and a candidate becomes a solution only after it is shown to satisfy the original equation. A step that can lose solutions, such as dividing both sides by an expression that might be zero or replacing $\ln\left(x^2\right)$ with $2 \ln x$ , breaks this guarantee and should be avoided.

In plain words The algebra hands you suspects. Each one still has to pass a check before it counts.

Example Solve $e^{2x} - e^x - 6 = 0$ . Put $u = e^x$ , so $e^{2x} = u^2$ and the equation becomes

\begin{gathered} u^2 - u - 6 = 0, \\[0.4em] \text{so} \quad (u - 3)(u + 2) = 0. \end{gathered}

Every solution $x$ of the original gives a root $u = e^x$ of this quadratic, so the candidates for $u$ are $3$ and $-2$ . Since $u = e^x$ must be positive, $u = -2$ is rejected. From $e^x = 3$ the single solution is $x = \ln 3$ . Checking, $e^{2 \ln 3} - e^{\ln 3} - 6 = 9 - 3 - 6 = 0$ .

Checking the original equation

Definition To check a candidate, substitute it into the original equation, as it was first written, and confirm that every expression is defined and that the two sides are equal. A check in a rewritten form is not enough, because the rewritten form may be defined where the original is not. When the solution is exact, check it exactly, not with a rounded decimal.

In plain words Go all the way back to the first line of the problem and test each candidate there.

Example Solve $\log_3 x + \log_3(x - 8) = 2$ . Combining gives $\log_3\left(x(x - 8)\right) = 2$ , so $x(x - 8) = 9$ and $x^2 - 8x - 9 = 0$ . The candidates are $x = 9$ and $x = -1$ .

At $x = 9$ the original reads $\log_3 9 + \log_3 1 = 2 + 0 = 2$ , so it is a solution. At $x = -1$ the combined form gives $\log_3 9 = 2$ , which looks fine, but the original contains $\log_3(-1)$ , which is undefined. So $x = -1$ is rejected and the only solution is $x = 9$ .

Check yourself

Solve $9^x = 27^{x - 1}$ with a common base.

Show the answer

Write both sides with base $3$ .

\begin{gathered} 3^{2x} = 3^{3(x - 1)}, \\[0.4em] \text{so} \quad 2x = 3x - 3 \quad \text{and} \quad x = 3. \end{gathered}

Check with $9^3 = 729$ and $27^2 = 729$ .

Solve $4e^{3x} = 10$ . Give the exact answer and a decimal approximation.

Show the answer

Isolating the exponential gives $e^{3x} = \frac{5}{2}$ , which is positive, so taking natural logarithms is allowed.

\begin{gathered} 3x = \ln \frac{5}{2}, \\[0.4em] \text{so} \quad x = \frac{1}{3} \ln \frac{5}{2} \approx 0.305. \end{gathered}

Solve $\log_2 x + \log_2(x + 2) = 3$ .

Show the answer

The original terms require $x > 0$ . Combining gives $x(x + 2) = 2^3 = 8$ , so $x^2 + 2x - 8 = 0$ and the candidates are $x = 2$ and $x = -4$ . At $x = 2$ the original reads $\log_2 2 + \log_2 4 = 1 + 2 = 3$ . The candidate $x = -4$ makes $\log_2(-4)$ undefined, so the only solution is $x = 2$ .

Solving Exponential and Logarithmic Equations

Key terms

Exponential equation

One-to-one exponential

Common base

Taking logarithms

Logarithmic equation

Argument restriction

Candidate solution

Checking the original equation

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus