4.4 Beyond the Basics
Introduction
In this section, we will introduce three types of equations that are just a little bit different than what we've seen before.
These new equations can easily be confused for one of the earlier types, so we will have to proceed with caution to identify them correctly.
Equations with Negative Coefficients
When solving an equation for a variable like \(x\), we sometimes run into the expression \(-x\). Since \(-x\) can be viewed as \(-1\cdot x\), if we are able to isolate \(-x\), then both sides could be divide by \(-1\) to give us the value for positive \(x\).
When solving an equation where the variable is negative, there are a few common errors that should be avoided. The first error is simply copying the equation incorrectly. For instance, \(2-x=9\) is not the same as \(x-2=9\).
Solve \(2 - x = 9\) for \(x\).
Solution
\begin{align} 2 - x &= 9\ && \hint{\text{Original equation}} \\ -x &= 7 && \hint{\text{Subtract }2} \\ x &= -7 && \hint{\text{Divide by }-1} \end{align}
The second error is not following the proper order of operations. For example, it is very tempting to simplify \(5-2x\) as \(3x\), but that places subtraction before multiplication, which is not the correct order.
Solve \(5 - 2x = 21\) for \(x\).
Solution
\begin{align} 5 - 2x &= 21 && \hint{\text{Given equation}} \\ -2x &= 16 && \hint{\text{Subtract }5} \\ x &= -8 && \hint{\text{Divide by }-2} \end{align}
Some prefer to rewrite the left side of the equation as \(-2x+5\) to make the operations easier to see.
\begin{align} -2x + 5 &= 21 && \hint{\text{Rewritten equation}} \\ -2x &= 16 && \hint{\text{Subtract }5} \\ x &= -8 && \hint{\text{Divide by }-2} \end{align}Both produce the same result.
This same error is also possible in situations where distribution could be used. In the next example we will solve \(7-3(x-1)=34\). Note that the \(7\) and the \(-3\) cannot be combined since the \(-3\) is multiplying \((x-1)\) and \(7\) is not.
Solve \(7-3(x-1)=34\) for \(x\).
Solution
\begin{align} 7-3(x-1) &= 34 && \hint{\text{Given equation}} \\ -3(x-1) &= 27 && \hint{\text{Subtract }7} \\ x -1 &= -9 && \hint{\text{Divide by }-3} \\ x &= -8 && \hint{\text{Add }1} \end{align}
Or, if you prefer to use distribution:
\begin{align} 7-3(x-1) &= 34 && \hint{\text{Given equation}} \\ 7-3x+3 &= 34 && \hint{\text{Distribute }-3} \\ 10-3x &= 34 && \hint{\text{Combine constants}} \\ -3x &= 24 && \hint{\text{Subtract }10} \\ x &= -8 && \hint{\text{Divide by }-3} \end{align}Equations with a Variable in the Denominator
The next type of equation to watch out for is one where the variable is in the denominator. It's critical to recognize that dividing something by \(x\) is not the same as dividing \(x\) by something. For instance, \(\frac{5}{x}\) is not the same as \(\frac{x}{5}\).
Luckily, equations with a variable in the denominator can usually be converted into ones where the variable isn't in the denominator.
This is done by "clearing the fractions" by multiplying each term by the least common denominator (LCD). We did a similar step to simplify equations in section 4.1.
Solve \(\frac{5}{x} = 8\) for \(x\).
Solution
Since the denominators are \(x\) and \(1\), the LCD is \(x\).
\begin{align} \frac{5}{x} &= 8 && \hint{\text{Given equation}} \\ x\cdot \frac{5}{x} &= x\cdot 8 && \hint{\text{Multiply by }x} \\ 5 &= 8x && \hint{\text{Simplify}} \\ x &= \frac{5}{8} && \hint{\text{Divide by }8} \end{align}Equations where both sides are fractions are known as proportions. With proportions, the process of clearing the fractions can be viewed as cross multiplication.
To cross multiply fractions, multiply the top of each fraction with the bottom of the other fraction. This converts an equation like \(\frac{a}{b}=\frac{c}{d}\) into one of the form \(a\cdot d = b\cdot c\).
Solve \(\frac{10}{x} = \frac{2}{7}\) for \(x\).
Solution
\begin{align} \frac{10}{x} &= \frac{2}{7} && \hint{\text{Given equation}} \\ 10 \cdot 7 &= 2 \cdot x && \hint{\text{cross multiply to eliminate the denominator}} \\ 70 &= 2x && \hint{\text{Simplify}} \\ 35 &= x && \hint{\text{Divide by }2} \end{align}
Sometimes cross multiplication will lead to situations where distribution and/or combining like terms is necessary.
Solve \(\frac{3}{x+1} = \frac{4}{3x}\) for \(x\).
Solution
\begin{align} \frac{3}{x+1} &= \frac{4}{3x} && \hint{\text{Given equation}} \\ 9x &= 4(x+1) && \hint{\text{cross multiply to eliminate the denominator}} \\ 9x &= 4x + 4 && \hint{\text{Distribute }4} \\ 5x &= 4 && \hint{\text{Subtract }4x} \\ x &= \frac{4}{5} && \hint{\text{Divide by }5} \end{align}
Cross multiplication is a handy tool, but don't forget that it can only be used when both sides are single fractions.
If one side has two or more terms then we cannot use cross multiplication and need to fall back on the method of multiplying by a common denominator. That's what we'll encounter in the next example.
Solve \(\frac{4}{x} + 3 = \frac{23}{5x}\) for \(x\).
Solution
Since the left side is not a single fraction, we cannot cross multiply. Instead, we'll multiply by the common denominator, which is \(5x\).
\begin{align} \frac{4}{x} + 3 &= \frac{23}{5x} && \hint{\text{Given equation}} \\ 5x\left(\frac{4}{x} + 3\right) &= 5x \cdot \frac{23}{5x} && \hint{\text{Multiply both sides by } 5x} \\ 5x \cdot \frac{4}{x} + 5x\cdot 3 &= 5x \cdot \frac{23}{5x} && \hint{\text{On the left, distribute } 5x} \\ \frac{20x}{x} + 15x &= \frac{115x}{5x} && \hint{\text{Multiply}} \\ 20 + 15x &= 23 && \hint{\text{Simplify fractions}} \\ 15x &= -3 && \hint{\text{Subtract 20 from both sides}} \\ x &= \frac{-3}{15} && \hint{\text{Divide both sides by }15} \\ x &= \frac{-1}{5} && \hint{\text{Reduce fraction}} \end{align}Equations with a Variable in the Exponent
This final type of equation might look familiar, and that's part of the challenge. It takes a careful eye to notice that expressions like \(x^2\) and \(2^x\) are not the same. The first has a variable base with power of \(2\). The other has a base of \(2\) and a variable exponent.
When the variable appears in the exponent, we call it an exponential equation. Again, the key feature of an exponential equation is that the variable is in the exponent, while the base remains constant.
For each equation, determine whether it is an exponential equation. If it is exponential, identify the base.
a. \(x^3 = 27\)
b. \(3^x = 27\)
c. \(2^{x+1} = 16\)
d. \(x^2 + 5x = 14\)
e. \(4 \cdot 5^x = 80\)
Solution
a. Not exponential. The variable \(x\) is the base, not in the exponent.
b. Exponential equation with base \(3\). The variable \(x\) is in the exponent.
c. Exponential equation with base \(2\). The variable expression \(x+1\) is in the exponent.
d. Not exponential. This is a quadratic equation with the variable in the base positions.
e. Exponential equation with base \(5\). The coefficient \(4\) is not part of the base.
In exponential equations, the exponent is a variable, not a fixed number like \(2\) or \(3\), so we cannot use roots to solve them. Sometimes, however, we can use our background knowledge of numbers to try a guess-and-check method of solving.
a. Solve \(7^x = 49\)
b. Solve \(10^x = 1000\)
c. Solve \(3^x = 81\)
d. Solve \(5^x = 125\)
Solution
Since the variables are in the exponent, these are all exponential equation. We will use guessing and checking to find the solutions.
a. \(x = 2\) because \(7^2 = 49\).
b. \(x = 3\) because \(10^3 = 1000\).
c. \(x = 4\) because \(3^4 = 81\).
d. \(x = 3\) because \(5^3 = 125\).
Logarithms
When we can't guess-and-check the solution to an exponential equation, an inverse operation called a logarithm can be used.
Just as subtraction undoes addition and division undoes multiplication, logarithms undo exponentiation. Logarithms allow you to find the exponent if you know what the base is.
The base of a logarithm is always written as a subscript. For example in the expression \(\log_2(16)\), the base is the number \(2\). And when no subscript is shown, like \(\log(100)\), the base is assumed to be \(10\).
The value of a logarithm is the exponent you need to put on the base in order to get the number in the parenthesis. With this in mind, it makes sense that \(\log_2(8)=3\) since \(2^3=8\), for example, or that \(\log(100)=2\) since \(10^2=100\).
A logarithm is the missing exponent. \[\log_b(y)=\square \quad \text{ only if } \quad b^{\ \square}=y\]
Use your understanding of what a logarithm is to find the following values mentally.
a. \(\log_7(49)=\square\)
b. \(\log(1000)=\square\)
c. \(\log_3(81)=\square\)
d. \(\log_5(125)=\square\)
Solution
For each logarithm, we need to figure out which exponent to put on the base to get the number in parentheses. That exponent will be our answer.
a. \(\log_7(49)=2\) because \(7^2 = 49\)
b. \(\log(1000)=3\) because \(10^3 = 1000\)
c. \(\log_3(81)=4\) because \(3^4 = 81\)
d. \(\log_5(125)=3\) because \(5^3 = 125\)
You might recognize that these are the same exact answers we had in Example 9, which shows that logarithms are a tool for solving exponential equations. Together, these examples illustrate the basic connection between logarithms and exponents: a logarithmic equation of the form \(\log_b(y) = x\) is equivalent to the exponential equation \(b^x = y\).
While the logarithms in this example were ones we were able to sort out mentally from our knowledge of exponents, finding the value of most logarithms is not a trivial exercise. Thankfully, all scientific calculators have a LOG button that finds logarithms when the base is \(10\). For example, if you enter LOG(100) your calculator will return 2 since \(10^2 = 100\).
If possible, use the LOG button on your calculator evaluate each expression. If needed, round to \(3\) decimal places and check your answers.
a. \(\log(200)\)
b. \(\log(649)\)
c. \(\log(937)\)
d. \(\log(-100)\)
Solution
a. \(\log(200)\approx 2.301\) which we check by seeing \(10^{2.301}=199.986\approx200\)
b. \(\log(649)\approx 2.812\) which we check by seeing \(10^{2.812}=648.634\approx649\)
c. \(\log(937)\approx 2.972\) which we check by seeing \(10^{2.972}=937.562\approx937\)
d. \(\log(-100)\) does not exist since no power of \(10\) gives a negative number. We can only put positive numbers into logarithms.
If you remember that \(\log\) of a number is the exponent you need to put on a \(10\) to get that number, you can see how it helps us solve equations like \(10^x = 100\). In general, if your equation is \(10^x=y\) then the solution is \(x=\log(y)\).
Keep in mind that the LOG button does not allow you to solve equations like \(3^x = 24\) directly since the base is \(3\), not \(10\). Methods for solving exponential equations like that are covered in a precalculus course. For now we'll stick to equations where the base is \(10\).
If possible, use the LOG button on your calculator to solve each equation. If needed, round to \(3\) decimal places.
a. Solve \(10^x = 80\)
b. Solve \(10^x = 5000\)
c. Solve \(2^x = 720\)
d. Solve \(10^x = 0.0242\)
Solution
a. \(x = \log(80) \approx 1.903\)
b. \(x = \log(5000) \approx 3.699\)
c. Cannot be solved directly with \(\log\) since the base is not \(10\).
d. \(x = \log(0.0242) \approx -1.616\)
The negative answer in Example 12(d) might surprise you after the comment in Example 11(d). It's perfectly fine for a negative number to come out of a logarithm, we just cannot put negative numbers into a logarithm. In fact, whenever the input is less than \(1\), the logarithm has to be negative because we need a negative exponent to make \(10^x\) small.
Conclusion
In this section we've expanded our equation-solving toolkit to handle several challenging variations: equations with negative coefficients, variables in denominators, and exponential equations. Being able to recognize the specific type of equation is critical so we can choose the appropriate solving technique.
While we've developed algebraic methods for solving many different types of equations, there's another powerful approach we haven't explored yet: solving equations graphically. In Chapter 5, we'll discover how graphs can help us visualize equations and their solutions.