Solving Exponential Equations, precalculus lesson

A town of 20000 people grows by 3 percent a year. When will it reach 30000? The question turns into the equation $20000(1.03)^t = 30000$ , where the unknown sits in the exponent and ordinary algebra cannot reach it. Logarithms can, because the power rule turns an exponent into a factor.

Key terms

Exponential equation

Definition An exponential equation is an equation in which the unknown appears in an exponent, such as $2^x = 10$ or $3^{x + 1} = 5^x$ . A solution is a real number at which both sides are defined and equal. The solutions of $f(x) = g(x)$ are exactly the first coordinates of the points where the graphs of $y = f(x)$ and $y = g(x)$ meet. For a base $b > 0$ with $b \ne 1$ , the power $b^x$ takes every positive value exactly once and never takes the value $0$ or a negative value.

In plain words You are looking for an exponent. A sketch of the two graphs tells you how many solutions to expect and roughly where they are.

Example The equation $2^x = 10$ asks where the rising curve $y = 2^x$ meets the horizontal line $y = 10$ . Since $2^3 = 8$ and $2^4 = 16$ , the crossing lies between $x = 3$ and $x = 4$ , and the exact answer is $x = \log_2 10 \approx 3.32$ . The equation $2^x = -4$ has no solution at all, because the curve stays above the $x$ -axis.

Common base

Definition Let $b > 0$ with $b \ne 1$ . The function $x \mapsto b^x$ is one-to-one, so for all real numbers $u$ and $v$

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

When both sides of an equation can be written as powers of the same valid base, the equation is therefore equivalent to the equation between the exponents. The base must really be valid. With base $1$ the rule fails, since $1^2 = 1^5$ .

In plain words If both sides are powers of the same number, set the exponents equal. No logarithms are needed and the answers come out exact.

Example Solve $4^{x + 1} = 8^x$ . Both $4 = 2^2$ and $8 = 2^3$ are powers of 2, so

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

As a check, $4^3 = 64$ and $8^2 = 64$ . Matching exponents can also lead to a quadratic. The equation $3^{x^2 - 2x} = 27 = 3^3$ gives $x^2 - 2x - 3 = 0$ , so $x = 3$ or $x = -1$ . Both work, since the exponent $x^2 - 2x$ equals $3$ at each of them.

Isolating the exponential

Definition To isolate the exponential means to use equivalent steps, such as adding the same number to both sides or dividing both sides by the same nonzero number, until the equation has the form

b^{g(x)} = c

with the power alone on one side, where the base satisfies $b > 0$ and $b \ne 1$ . If $c \le 0$ , the equation has no solution, because $b^{g(x)} > 0$ wherever it is defined. If $c > 0$ , it is equivalent to $g(x) = \log_b c$ .

In plain words Peel away everything around the power before you take a logarithm. The logarithm of a sum cannot be split, so taking logarithms too early leads nowhere.

Example Solve $5e^{2x} - 4 = 11$ . Adding 4 and dividing by 5 gives $e^{2x} = 3$ . Then $2x = \ln 3$ , so

x = \frac{\ln 3}{2} \approx 0.5493.

Taking $\ln$ of both sides at the start would give $\ln\left(5e^{2x} - 4\right) = \ln 11$ , which no rule simplifies. Isolating also exposes equations with no solution. The equation $3 \cdot 2^x + 7 = 1$ leads to $2^x = -2$ , which is impossible.

Applying the natural logarithm

Definition For positive numbers $A$ and $B$ ,

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

because $\ln$ is one-to-one on $(0, \infty)$ . So taking $\ln$ of both sides of an equation whose two sides are positive is an equivalent step. Together with the power rule $\ln\left(b^u\right) = u \ln b$ for $b > 0$ , it turns an unknown exponent into an unknown factor. Any valid base would work. The natural logarithm is the usual choice because calculators have it and calculus prefers it.

In plain words When the bases do not match, take $\ln$ of both sides and pull the exponents down. What remains is an ordinary linear or polynomial equation.

Example For the town, dividing $20000(1.03)^t = 30000$ by 20000 gives $1.03^t = 1.5$ . Taking $\ln$ of both sides,

\begin{gathered} t \ln 1.03 = \ln 1.5, \\[0.4em] t = \frac{\ln 1.5}{\ln 1.03} \approx 13.72, \end{gathered}

so the town reaches 30000 people after about 13.7 years. Two different bases are handled the same way. From $3^{x + 1} = 5^x$ you get $(x + 1) \ln 3 = x \ln 5$ , so $x(\ln 5 - \ln 3) = \ln 3$ and

x = \frac{\ln 3}{\ln 5 - \ln 3} \approx 2.1507.

Exact form

Definition An answer is in exact form when it is written with integers, fractions, radicals, logarithms, $e$ and similar symbols, with no rounding. Exact forms are not unique, and two of them name the same number when the rules of logarithms turn one into the other. A decimal answer is an approximation and is written with $\approx$ .

In plain words Give the exact answer first and the decimal second. The exact answer is what you can check with algebra, and the decimal is what you report in an application.

Example Solving $e^{0.04t} = 3$ gives

t = \frac{\ln 3}{0.04} = 25 \ln 3 \approx 27.47.

The exact form checks perfectly, since $e^{0.04 \cdot 25 \ln 3} = e^{\ln 3} = 3$ . The solution of $3^{x + 1} = 5^x$ above has several exact forms,

x = \frac{\ln 3}{\ln 5 - \ln 3} = \frac{\ln 3}{\ln \frac{5}{3}} = \log_{5/3} 3,

and all three name the number $2.15066\ldots$ , at which both sides of the equation equal about $31.86$ .

Equations of quadratic type

Definition An equation such as $e^{2x} - e^x - 6 = 0$ is of quadratic type in $e^x$ , because $e^{2x} = \left(e^x\right)^2$ . The substitution $u = e^x$ turns it into a quadratic equation in $u$ . Each real root $u$ then gives the equation $e^x = u$ , which has the single solution $x = \ln u$ when $u > 0$ and no solution when $u \le 0$ . The same works for $b^{2x} = \left(b^x\right)^2$ with any valid base $b$ .

In plain words Spot the hidden square, solve the quadratic, then go back from $u$ to $x$ . A root $u$ that is negative or zero gives nothing, because powers of a positive base are positive.

Example With $u = e^x$ the equation $e^{2x} - e^x - 6 = 0$ becomes $u^2 - u - 6 = 0$ , so $(u - 3)(u + 2) = 0$ . The root $u = 3$ gives $x = \ln 3 \approx 1.0986$ . The root $u = -2$ would need $e^x = -2$ , which never happens. So the only solution is $x = \ln 3$ , and indeed

e^{2 \ln 3} - e^{\ln 3} - 6 = 9 - 3 - 6 = 0.

Check yourself

Solve $9^x = 27^{x - 1}$ .

Show the answer

Write both sides as powers of 3. Then $3^{2x} = 3^{3x - 3}$ , so $2x = 3x - 3$ and $x = 3$ . As a check, $9^3 = 729 = 27^2$ .

Solve $4e^{3x} - 7 = 13$ . Give the exact answer and a decimal.

Show the answer

Isolating the power gives $e^{3x} = 5$ , so $3x = \ln 5$ and

x = \frac{\ln 5}{3} \approx 0.5365.

Solve $2^x = 3^{x - 2}$ .

Show the answer

Taking $\ln$ of both sides gives $x \ln 2 = (x - 2) \ln 3$ , so $x(\ln 3 - \ln 2) = 2 \ln 3$ and

x = \frac{2 \ln 3}{\ln 3 - \ln 2} = \frac{\ln 9}{\ln 1.5} \approx 5.4190.

At this value both sides are about $42.78$ .

Solving Exponential Equations

Key terms

Exponential equation

Common base

Isolating the exponential

Applying the natural logarithm

Exact form

Equations of quadratic type

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus