MA001 Study Guide

8a. apply properties of logarithms to rewrite and simplify logarithmic expressions, including the product, quotient, and power rules and the change-of-base formula

• Explain how to use logarithmic properties to rewrite the logarithm of a sum, the logarithm of a product, and the logarithm of a power.
• Describe how the power rule can simplify $\log _{b} 1, \log _{b} b$, and $\log _{b} b^{n}$.
• Rewrite a logarithm with a base other than $10$ or $e$ using the change of base formula. How is this formula helpful in using a calculator to evaluate a logarithm?
• Given an expression with multiple terms containing logarithms, use the logarithmic rules to rewrite the expression as a single logarithm.

Before the widespread availability of scientific and graphing calculators, complicated calculations were often completed using tables of logarithms by converting multiplications to additions, divisions to subtractions, and powers to basic multiplications by using the following properties of logarithms for bases $b > 0, b \neq 1$.

• Product rule for logarithms: $\log _{b}(M \cdot N)=\log _{b} M+\log _{b} N$
• Quotient rule for logarithms: $\log _{b} \frac{M}{N}=\log _{b} M-\log _{b} N$
• Power rule for logarithms: $\log _{b} M^{N}=N \cdot \log _{b} M$

Consider the expression $\log \frac{x y^{4}}{z}$ for $x, y$, and $z$ positive real numbers. Using these three logarithmic rules, this expression can be rewritten as $\log \frac{x y^{4}}{z}=\log x+4 \cdot \log y-\log z$. These rules can also be applied in reverse to rewrite an expression such as $\log 5-\log 7+3 \log x$ as $\log \frac{5 x^{3}}{7}$.

These rules can be used, together with the related exponential equation, to evaluate three special cases.

• $\log _{b} 1=0$
• $\log _{b} b=1$
• $\log _{b} b^{n}=n$

Scientific and graphing calculators have special keys for common logarithms (base 10) and natural logarithms (base e). Sometimes, however, there is a need to evaluate a logarithm with a different base. The change-ofbase formula can be used to rewrite a logarithm from one base to a quotient of logarithms with another base.

• Change-of-base formula for logarithms: For $M > 0, n > 0, b > 0, n \neq 1, b \neq 1, \log _{b} M=\frac{\log _{n} M}{\log _{n} b}$.

So, if the base $b$ is not 10 or $e$, then choose $n$ to be 10 or $e$ so that the logarithm can be evaluated using a calculator. For instance, $\log _{4} 21=\frac{\log 21}{\log 4} \approx \frac{1.322}{0.602} \approx 2.196$. This result makes sense because $y=\log _{4} 21$ is equivalent to $4^{y}=21$, so $2 < y < 3$. Also, evaluating $4^{2.196} \approx 20.995$.

Review material in Using the Definition of a Logarithm to Solve Logarithmic Equations.

8b. solve exponential equations using like bases and logarithms

• Explain how exponential equations can be solved if the exponential expressions on both sides of the equation have the same base. How can this technique be used even if the exponential expressions do not have the same base?
• How can logarithms be used to solve exponential equations? What are extraneous solutions?

Some types of exponential equations can be solved using known algebraic techniques. One such instance occurs when two algebraic expressions have the same base; in that case, the exponents are set equal to each other.

• For $b > 0$ and $b \neq 1, b^{M}=b^{N}$ if and only if $M=N$.

Sometimes the bases are already written in a form that indicates they are equal. For instance, if $3^{2 x-5}=$ $3^{x+7}$, then $2 x-5=x+7$, so $x=12$. However, in other cases, the bases are not the same but can be rewritten so they are the same. Consider the equation $3^{x-4}=81^{x+1}$. Although the bases are not currently the same, 81 is a power of 3, so rewrite $81^{x+1}$ as $\left(3^{4}\right)^{x+1}=3^{4 x+4}$. Now, $3^{x-4}=81^{x+1}$ can be rewritten as $3^{x-4}=3^{4 x+4}$ so $x-4=4 x+4$ or $x=-\frac{8}{3}$.

At times, however, there is no way to rewrite expressions so the exponential expressions have the same base. In that case, an appropriate technique is to take the logarithm of both sides of the equation, applying the rules from objective $8 \mathrm{a}$ as needed. Because calculators have dedicated keys for common logarithms and natural logarithms, usually base 10 or base $e$ should be used as the base of the logarithm to be applied to each side of the equation. To solve $6^{x+2}=7^{3 x}$, take the common logarithm of both sides and then apply the rules of logarithms from objective 8a.

$\log \left(6^{x+2}\right)=\log \left(7^{3 x}\right)$

$(x+2) \log 6=(3 x) \log 7$

$2 \log 6=x(3 \log 7-\log 6)$

$x=\frac{2 \log 6}{3 \log 7-\log 6} \approx \frac{1.5563}{2.5353-0.7782} \approx 0.8857$

Whenever the logarithm of both sides is taken, it is essential to check the solution to ensure that none of the potential solutions are extraneous, that is, a solution obtained during the algebraic process that does not solve the original equation. In this case, $6^{0.8857+2}=7^{3 \cdot 0.8857}$, or $175.9993 \approx 175.9989$, close enough given potential rounding errors.

Review material in Solving Exponential Equations Using Logarithms.

8c. solve logarithmic equations using the definition of a logarithm and the one-to-one property of logarithms

• Explain how to use the definition of a logarithm, together with the rules of logarithms, to solve a logarithmic equation.
• Explain how the one-to-one property of logarithms can be used to solve logarithmic equations. Compare the use of this property to the use of the same base property when solving exponential equations.

Two techniques are often helpful in solving logarithmic equations. The first uses the fact that the logarithmic equation $y=\log _{b} x$ is equivalent to $b^{y}=x$. So, start by using the rules of logarithms to rewrite multiple logarithmic terms as a single logarithm and then rewrite the result in exponential form and solve.

Consider the equation $\log (4 x+4)+\log (x-14)=3$. Because the left side of the equation is the sum of two logarithms, it can be rewritten as the logarithm of a product: $\log [(4 x+4)(x-14)]$. Then rewrite the logarithmic equation into exponential form:

$\log [(4 x+4)(x-14)]=3$ is equivalent to $(4 x+4)(x-14)=10^{3}$

So, $4 x^{2}-52 x-56=1000$, or $x^{2}-13 x-264=0$. Factoring or using the quadratic formula leads to $x=24$ or $x=-11$, but $x=-11$ is an extraneous solution because it is not possible to take the logarithm of a negative number. Checking the solution $x=24$ results in $\log 100+\log 10=2+1=3$, so it checks.

The technique used to solve an exponential equation if two exponential expressions have the same base has an analogy that can be used to solve a logarithmic equation if two logarithmic expressions have the same base.

• For $b > 0, b \neq 1$ and $M, N > 0, \log _{b} M=\log _{b} N$ if and only if $M=N$.

Consider the equation $\log (6 x+8)=\log (2 x+28)$. Both logarithmic expressions have base 10, so $6 x+8=2 x+28$, or $x=5$. Checking this value in the equation leads to $\log 38=\log 38$

Review material in Solving Exponential Equations Using Logarithms.

8d. solve applications of exponential and logarithmic equations, including half-life and Newton's law of cooling

• Write the formula for compound interest and describe the meanings of the component parts of the formula. 
• Compare the compound interest formula with the formula for continuous growth or decay.
• Identify the exponential equation used for half-life and identify the meaning of all parts of the equation.
• Identify an equation for doubling time.
• Be able to write and use the formula for Newton's law of cooling.

Exponential and logarithmic functions appear in many real-world situations. Solving real-world applications uses the same techniques as described in objectives 8b and 8c. But some applications appear so commonly that they deserve special attention.

One such application is compound interest, determined by the equation $A(t)=P\left(1+\frac{r}{n}\right)^{n t}$, with $A(t)$ representing the amount in the account after $t$ years, $P$ representing the initial principal or amount, $r$ representing the decimal for the interest rate, and $n$ representing the compounding periods per year (i.e., 2 for semiannually, 4 for quarterly, 12 for monthly). Suppose $\$ 100$ is deposited into an account at an interest rate of $3 \%$ compounded twice a year. After 5 years, the amount in the account would be $100\left(1+\frac{0.03}{2}\right)^{2 \cdot 5} \approx \$ 107.73$.

Many situations modify the compound interest formula to be continuous growth or decay by using $e$. In this case, the equation becomes $A(t)=A_{0} e^{k t}$, where $A(t)$ represents the amount after time $t, A_{0}$ is the initial value, and $k$ is the growth factor per unit time. If $k > 0$, the situation models continuous growth; if $k < 0$, the situation models continuous decay. For example, assume a population of 1000 bacteria decreases by $3 \%$ every 5 hours. To find the number of bacteria that exist after 20 hours, first recognize that 20 hours is equivalent to 4 of the 5 hour time periods over which the $3 \%$ applies. Then the number of bacteria left is $1000 e^{-0.03 \cdot 4} \approx 887$ bacteria.

An important scientific application using exponential equations is that of half-life, or the time period in which half of a radioactive substance decays. An equation for half-life is $A(t)=A_{0}\left(\frac{1}{2}\right)^{t / T}$, where $A_{0}$ represents the initial amount, $t$ represents the time period of interest, $T$ represents the half-life, and $A(t)$ represents the amount remaining after time $t$. Notice that the exponent $\frac{t}{T}$ represents how many half-lives occur in the $t$ time period. For instance, the half-life of gallium-67 is about 80 hours. Then after 240 hours, an initial quantity of 10 micrograms would decay so that only $10\left(\frac{1}{2}\right)^{240 / 80} \approx 1.25$ micrograms.

In contrast to half-life, some applications are interested in doubling time, that is, the amount of time it takes for a quantity to double. The equation for doubling time $t$ is $t=\frac{\ln 2}{k}$, where $k$ is the growth factor per unit time. So, if a quantity has a growth factor of $6 \%$ per year, then it will double in $\frac{\ln 2}{0.06} \approx 11.55$ years.

Newton's law of cooling provides insight into how quickly a hot object will cool to approximate the surrounding temperature:

• $T(t)=A e^{k t}+T_{s}$, where $T(t)$ is the temperature after time $t, A$ is the difference between the initial temperature of an object and the surrounding temperature, $k$ is the cooling rate, and $T_{s}$ is the temperature of the surrounding air.

Such an equation is solved using the same techniques for solving exponential equations as previously reviewed.

Suppose a boiling pot of soup at a temperature of $100^{\circ}$ is placed on a table in a room whose temperature is $72^{\circ}$. If the temperature of the soup is $90^{\circ}$ after 5 minutes, what equation can be used to find the temperature for any time $t$? In this case, $A=100-72=28$ and $T_{s}=72$. Because $T(5)=90$, the equation becomes $90=$ $28 e^{5 k}+72$, so $e^{5 k}=\frac{90-72}{28}$ leading to $k \approx-0.088$. So, the equation becomes $T(t)=28 e^{-0.088 t}+72$.

Review material in Models of Exponential Growth and Decay, Solving Applied Problems Using Exponential and Logarithmic Equations, and Use Data to Build a Logarithmic Model.

8e. differentiate between exponential growth and decay

• Contrast exponential growth with exponential decay.

The equation $y=A_{0} e^{k t}$, where $A_{0}$ is the initial value, $k$ is the growth factor, and $t$ is the time, represents an exponential model that characterizes growth if $k > 0$ or decay if $k < 0$. An exponential growth function is always increasing; in contrast, an exponential decay function is always decreasing. For both exponential growth and decay, the domain is $(-\infty, \infty)$ and the range is $(0, \infty)$. Both functions have a $y$-intercept at $\left(0, A_{0}\right)$ and a horizontal asymptote at $y=0$. Neither function has a $x$-intercept. The graphs of an exponential growth and exponential decay are similar to the graphs in objective 7c.

Examples of exponential growth are compound interest and population growth that is a percentage over time. Examples of exponential decay are half-life and Newton's law of cooling.

$f(x)=\frac{c}{1+a e^{-b x}}$

In this equation, $c$ is the carrying capacity, $\frac{c}{1+a}$ is the initial value, and $b$ is determined by the growth factor; because $\frac{c}{1+a}$ is the value of $f(0)$, knowing the value of the intercept enables the value of the coefficient a to be determined.

Consider the following problem.

• A middle school has a student population of 1500 students. Four students hear a rumor; 9 hours later, 32 students have heard the rumor. What is an equation for the number of students who will have heard the rumor after $t$ hours?

It makes sense to model this problem with a logistic growth model because the maximum number of students who can have heard the rumor is 1500; this means $c=1500$. The two points on the curve $(0,4)$ and $(9,32)$ can be used to find an equation. The point $(0,4)$ leads to $4=\frac{1500}{1+a}$ or $a=374$. Then, use the point $(9,32)$ to obtain $32=\frac{1500}{1+374 \cdot e^{-9 b}}$ which leads to $e^{-9 b}=\frac{1468}{11968}$ so $b=\frac{\ln (1468 / 11968)}{-9} \approx 0.233$. Thus, the equation is $f(x)=$ $\frac{1500}{1+374 e^{-0.233 x}}$.

Review material in Use Data to Build a Logarithmic Model.

8f. identify the key components of a logistic growth model

• Describe an equation for a logistic model. How does a logistic model differ from an exponential model?
• What is the carrying capacity of a logistic model?

An exponential growth model suggests that growth continues forever as indicated by the fact that an exponential function increases without bound as the input value increases without bound. But in the real world, growth can generally not continue in this manner forever but reaches a limiting value and then begins to slow. So, many real-world situations involving exponential growth would be better modeled by a logistic growth function, which is exponential initially but decreases as the input reaches an upper bound or limiting value known as the carrying capacity. The equation for a logistic growth model is

$f(x)=\frac{c}{1+a e^{-b x}}$

In this equation, $c$ is the carrying capacity, $\frac{c}{1+a}$ is the initial value, and $b$ is determined by the growth factor; because $\frac{c}{1+a}$ is the value of $f(0)$, knowing the value of the intercept enables the value of the coefficient $a$ to be determined.

Consider the following problem.

• A middle school has a student population of 1500 students. Four students hear a rumor; 9 hours later, 32 students have heard the rumor. What is an equation for the number of students who will have heard the rumor after $t$ hours?

It makes sense to model this problem with a logistic growth model because the maximum number of students who can have heard the rumor is 1500 ; this means $c=1500$. The two points on the curve $(0,4)$ and $(9,32)$ can be used to find an equation. The point $(0,4)$ leads to $4=\frac{1500}{1+a}$ or $a=374$. Then, use the point (9, 32) to obtain $32=\frac{1500}{1+374 \cdot e^{-9 b}}$ which leads to $e^{-9 b}=\frac{1468}{11968}$ so $b=\frac{\ln (1468 / 11968)}{-9} \approx 0.233$. Thus, the equation is $f(x)=$ $\frac{1500}{1+374 e^{-0.233 x}}$.

Review material in Use Data to Build a Logarithmic Model.

8g. build exponential and logarithmic regression models from data, and assess their fit

• Given a set of data that appears to be exponential, find an appropriate exponential regression equation. Use the equation to predict the output for other input values.
• Given a set of data that appears to be logarithmic, find an appropriate regression equation. Use the equation to predict the output for other input values.

In objective $4 \mathrm{~h}$, there were procedures for finding a linear regression to fit a set of data, often using a graphing utility. The correlation coefficient $r$ provides insights into how well the model fits the data; the closer $r$ is to 1 or 1, the better the model fits the data. Similar procedures are used to fit regression models where the data appear to grow or decay exponentially or to fit a logarithmic function. Most graphing calculators or comparable software utilities have options for an exponential regression (ExpReg) or natural logarithmic regression ($\mathrm{LnReg}$); once again, the value of $r$ provides insight into how well the equation fits the data, with better models having values of $r$ closer to 1 or $-1$. After a regression equation is found, it can be used to predict new output values by substituting input values into the equation.

Consider the data in the table below and the graphs of those data values. The data for function $f$ appear to be exponential while that for function $g$ appear to be logarithmic.

When the data are entered into a graphing calculator, the following regression equations are obtained.

• $f(x)=0.54 \cdot 2.95^{x}$ with $r=0.9999$
• $\mathrm{g}(x)=7.94-1.25 \cdot \ln x$ with $r=-0.996$

These equations can be used to predict the output for other input values. For instance, $f(10) \approx 26,953$ and $g(10) \approx 5.062.$

Review material in Models of Exponential Growth and Decay.

Unit 8 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• carrying capacity
• change-of-base formula for logarithms
• compound interest formula
• continuous growth or decay
• doubling time
• extraneous solution
• half-life
• logistic growth model
• Newton's law of cooling
• power rule for logarithms
• product rule for logarithms
• quotient rule for logarithms