Skills Review for First-order Linear Equations

Learning Outcomes

Solve rational equations by clearing denominators
Apply the inverse property to simplify logarithms
Define and use the power rule for logarithms to rewrite expressions
Write function equations using given conditions

In the First-order Linear Equations section, we will explore how to solve yet another special type of differential equation. Here we will review how to clear fractions from and manipulate rational equations.

Manipulate Rational Equations

Equations that contain rational expressions are called rational equations. For example, $\frac{2y+1}{4}=\frac{x}{3}$ is a rational equation (of two variables).

One of the most straightforward ways to solve a rational equation for the indicated variable is to eliminate denominators with the common denominator and then use properties of equality to isolate the indicated variable.

Solve for $y$ in the equation $\frac{1}{2}y-3=2-\frac{3}{4}x$ by clearing the fractions in the equation first.

Multiply both sides of the equation by $4$ , the common denominator of the fractional coefficients.

$\begin{array}{c}\frac{1}{2}y-3=2-\frac{3}{4}x\\ 4\left(\frac{1}{2}y-3\right)=4\left(2-\frac{3}{4}x\right)\\\text{}\,\,\,\,4\left(\frac{1}{2}y\right)-4\left(3\right)=4\left(2\right)+4\left(-\frac{3}{4}x\right)\\2y-12=8-3x\\\underline{+12}\,\,\,\,\,\,\underline{+12}\\ 2y=20-3x\\ \frac{2y}{2}=\frac{20-3x}{2} \\ y=10-\frac{3}{2}x\end{array}$

Example: Manipulating a Rational Equation

Solve the equation $\frac{x+5}{8}=\frac{7}{y}$ for $y$ .

Show Solution

Find the least common denominator of $8$ and $y$ . $8y$ will be the LCD.

Multiply both sides of the equation by the common denominator, $8y$ , to keep the equation balanced and to eliminate the denominators.

$\begin{array}{r}8y\cdot \frac{x+5}{8}=\frac{7}{y}\cdot 8y\,\,\,\,\,\,\,\\\\\frac{8y(x+5)}{8}=\frac{7(8y)}{y}\,\,\,\,\,\,\\\\\frac{8y}{8}\cdot (x+5)=\frac{7(8y)}{y}\\\\\frac{8}{8}\cdot y(x+5)=7\cdot 8\cdot \frac{y}{y}\\\\1\cdot y(x+5)=56\cdot 1\,\,\,\\\\ y(x+5)=56\,\,\,\,\,\,\,\, \\\\ \frac{y(x+5)}{x+5}=\frac{56}{x+5}\,\,\,\,\,\,\,\,\, \\\\ y=\frac{56}{x+5}\,\,\,\,\,\,\,\,\,\,\,\end{array}$

In the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they do not share any common factors.

Example: Manipulating a Rational Equation

Solve the equation $x=\frac{4}{3y+1}$ for $y$ .

Show Solution

Clear the denominators by multiplying each side by the common denominator. The common denominator is $3y+1$ .

$\begin{array}{c}\left(3y+1\right)\left(x\right)=\left(\frac{4}{3y+1}\right)\left(3y+1\right)\end{array}$

Simplify common factors.

$\begin{array}{c}\left(3y+1\right)\left(x\right)=\left(\frac{4}{\cancel{3y+1}}\right)\left(\cancel{3y+1}\right)\\\left(3y+1\right)\left(x\right)=4\end{array}$

We can now use the addition and multiplication properties of equality to solve it.

$\begin{array}{c}\left(3y+1\right)\left(x\right)=4 \\ \\ \frac{(3y+1)(x)}{x}=\frac{4}{x} \\ \\ 3y+1=\frac{4}{x} \\ \,\,\,\,\,\,\,\,\,\,\underline{-1}\,\,\,\,\,\,\underline{-1} \\ 3y=\frac{4}{x}-1\\ \\ \frac{3y}{3}=\frac{4}{3x}-\frac{1}{3}\\ \\ y=\frac{4}{3}x-\frac{1}{3}\end{array}$

Apply the Inverse Property to Simplify Logarithms

Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.

$\begin{array}{l}{\mathrm{log}}_{b}1=0\\{\mathrm{log}}_{b}b=1\end{array}$

For example, ${\mathrm{log}}_{5}1=0$ since ${5}^{0}=1$ and ${\mathrm{log}}_{5}5=1$ since ${5}^{1}=5$ .

Next, we have the inverse property.

$\begin{array}{l}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}$

For example, to evaluate $\mathrm{log}\left(100\right)$ , we can rewrite the logarithm as ${\mathrm{log}}_{10}\left({10}^{2}\right)$ and then apply the inverse property ${\mathrm{log}}_{b}\left({b}^{x}\right)=x$ to get ${\mathrm{log}}_{10}\left({10}^{2}\right)=2$ .

To evaluate ${e}^{\mathrm{ln}\left(7\right)}$ , we can rewrite the logarithm as ${e}^{{\mathrm{log}}_{e}7}$ and then apply the inverse property ${b}^{{\mathrm{log}}_{b}x}=x$ to get ${e}^{{\mathrm{log}}_{e}7}=7$ .

Example: Simplify Logarithms

Simplify each of the following.

a) $e^{\ln(7x)}$

b) $\log_9(9^4)$

c) $5^{\log_5(3y)}$

Show Solution

a) Using the second inverse property, $e^{\ln(7x)}=7x$

b) Using the first inverse property, $\log_9(9^4)=4$

c) Using the second inverse property, $5^{\log_5(3y)}=3y$

Try It

Simplify each of the following.

a) $b^{\log_b(15x)}$

b) $\log_2(2^a)$

c) $e^{\ln(4y)}$

Show Solution

a) $15x$

b) $a$

c) $4y$

Use the Power Property for Logarithms

A General Note: The Power Rule for Logarithms

The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.

${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$

Example: Expanding a Logarithm with Powers

Rewrite ${\mathrm{log}}_{2}{x}^{5}$ .

Show Solution

The argument is already written as a power, so we identify the exponent, 5, and the base, x, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

${\mathrm{log}}_{2}\left({x}^{5}\right)=5{\mathrm{log}}_{2}x$

Try It

Rewrite $\mathrm{ln}{x}^{2}$ .

Show Solution

$2\mathrm{ln}x$

Example: Rewriting an Expression as a Power before Using the Power Rule

Rewrite ${\mathrm{log}}_{3}\left(25\right)$ using the power rule for logs.

Show Solution

Expressing the argument as a power, we get ${\mathrm{log}}_{3}\left(25\right)={\mathrm{log}}_{3}\left({5}^{2}\right)$ .

Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

${\mathrm{log}}_{3}\left({5}^{2}\right)=2{\mathrm{log}}_{3}\left(5\right)$

Try It

Rewrite $\mathrm{ln}\left(\frac{1}{{x}^{2}}\right)$ .

Show Solution

$-2\mathrm{ln}\left(x\right)$

Write Function Equations Using Given Conditions

(See Module 4, Skills Review for Basics of Differential Equations)

Differential Equations: Prerequisite Material