One way to graph a log function is to first make a table of values for its inverse function, the exponential function with the same base, then interchange the variables.
CheckpointA.10.1.
Complete the table of values and graph on the same grid: \(~f(x)=x^3~\) and \(~g(x)=3^x~\)
\(~~~x~~~\)
\(~~~0~~~\)
\(~~~1~~~\)
\(~~~2~~~\)
\(~~~3~~~\)
\(~~~4~~~\)
\(~~~5~~~\)
\(~~~6~~~\)
\(~~~8~~~\)
\(~~10~~~\)
\(~f(x)~\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(~g(x)~\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
Use your tables from part (a) to graph \(~h(x)=\sqrt[3]{x}~\) and \(~j(x)=\log_3 {(x)}~\) on the same grid.
Answer.
SubsubsectionA.10.1.2Use function notation
A log function is the inverse of the exponential function with the same base, and vice versa.
ExampleA.10.2.
For each function \(~f(x)~\text{,}\) decide whether \(f(a+b)=f(a)+f(b)\text{.}\)
\(\displaystyle f(x)=3^x\)
\(\displaystyle f(x)=\log_3 {(x)}\)
Solution.
\(f(a+b)=3^{a+b},~\) and \(~f(a)+f(b) = 3^a+3^b.~\)
But \(~3^{a+b}~\) is not equivalent to \(~3^a+3^b;~\) in fact \(~3^{a+b} =3^a \cdot 3^b.\)
So for this function, \(f(a+b) \not= f(a)+f(b)\text{.}\)
\(f(a+b)=\log_3 {(a+b)},~\) and \(~f(a)+f(b) = \log_3 {(a)} +\log_3 {(b)}.~\)
But \(~\log_3 {(a+b)}~\) is not equivalent to \(~\log_3 {(a)} +\log_3 {9};~\) in fact \(~\log_3 {(ab)} = \log_3 {(a)} +\log_3 {(b)}.\)
So for this function, \(f(a+b) \not= f(a)+f(b)\text{.}\)
CheckpointA.10.3.
\(g(x)=5^x~\text{.}\) Evaluate and simplify if possible.
\(\displaystyle g(3+t)\)
\(\displaystyle g(3t)\)
Answer.
\(\displaystyle 125 \cdot 5^{t}\)
\(\displaystyle 125^t\)
ExampleA.10.4.
\(q(x)=9^x~\) and \(p(x)\) is its inverse function. Evaluate if possible.
\(\displaystyle q\left(\dfrac{1}{2}\right)\)
\(\displaystyle p(3)\)
\(\displaystyle q(0)\)
\(\displaystyle p(0)\)
Solution.
\(\displaystyle 3\)
\(\displaystyle \dfrac{1}{2}\)
\(\displaystyle 1\)
undefined
CheckpointA.10.5.
\(h(x)=\log_4 {(x)}.~\) Evaluate if possible.
\(\displaystyle h(4)\)
\(g(4)\text{,}\) where \(g\) is the inverse function for \(h\)
\(\displaystyle h(0)\)
\(\displaystyle g(0)\)
Answer.
\(\displaystyle 1\)
\(\displaystyle 256\)
undefined
\(\displaystyle 1\)
CheckpointA.10.6.
\(f(x)=\log_8 {(x)}~\text{.}\) Evaluate and simplify if possible.
\(\displaystyle f(64p)\)
\(\displaystyle f(64+p)\)
Answer.
\(\displaystyle 2+\log_8 {(p)}\)
cannot be simplified
SubsubsectionA.10.1.3Use the properties of logarithms
The three properties of logarithms are helpful in making computations involving logs.
Properties of Logarithms.
If \(x\text{,}\)\(y\text{,}\)\(b \gt 0\text{,}\) and \(b\ne 1\text{,}\) then
Because \(\log {(x)}\) grows very slowly, we can use logs to compare quantities that vary greatly in magnitude.
ExampleA.10.12.
Complete the table. Round the values to one decimal place.
\(x\)
\(~1~\)
\(~5~\)
\(25\)
\(125\)
\(625\)
\(\log {(x)}\)
\(\)
\(\)
\(\)
\(\)
\(\)
Plot the values of \(x\) on a log scale.
Each time we multiply \(x\) by 5, how much does the logarithm increase? What is \(\log {(5)}\text{,}\) to one decimal place?
Solution.
\(x\)
\(~1~\)
\(~5~\)
\(25\)
\(125\)
\(625\)
\(\log x\)
\(0\)
\(0.7\)
\(1.4\)
\(2.1\)
\(2.8\)
Each time we multiply \(x\) by 5, the log of \(x\) increases by 0.7, because \(\log {(5)} = 0.7\text{.}\) This is an application of the log properties:
Complete the table. Round the values to one decimal place.
\(x\)
\(~5~\)
\(~10~\)
\(~20~\)
\(~40~\)
\(~80~\)
\(\log {(x)}\)
\(\)
\(\)
\(\)
\(\)
\(\)
Plot the values of \(x\) on a log scale.
Each time we multiply \(x\) by 2, how much does the logarithm increase? What is \(\log {(2)}\text{,}\) to one decimal place?
Answer.
\(x\)
\(~5~\)
\(~10~\)
\(~20~\)
\(~40~\)
\(~80~\)
\(\log {(x)}\)
\(0.7\)
\(1\)
\(1.3\)
\(1.6\)
\(1.9\)
\(\displaystyle 0.3;~0.3\)
CheckpointA.10.14.
Complete the table. Round the values to one decimal place.
\(x\)
\(0.25\)
\(~1~\)
\(~4~\)
\(~16~\)
\(~64~\)
\(256\)
\(\log {(x)}\)
\(\)
\(\)
\(\)
\(\)
\(\)
\(\)
Plot the values of \(x\) on a log scale.
Each time we multiply \(x\) by 4, how much does the logarithm increase? What is \(\log {(4)}\text{,}\) to one decimal place?
Answer.
\(x\)
\(0.25\)
\(~1~\)
\(~4~\)
\(~16~\)
\(~64~\)
\(256\)
\(\log {(x)}\)
\(-0.6\)
\(0\)
\(0.6\)
\(1.2\)
\(1.8\)
\(2.4\)
\(\displaystyle 0.6;~0.6\)
SubsubsectionA.10.2.2Compare quantities
There is often more than one way to express a comparison with mathematical notation.
ExampleA.10.15.
When we say that "\(A\) is 3 times larger than \(B\text{,}\)" we mean that \(A=3B\text{.}\)
ExampleA.10.16.
When we say that "\(A\) is 3 more than \(B\text{,}\)" we mean that \(A=B+3\text{.}\)
Use these equations for the following Checkpoints.
\(\displaystyle x=5H\)
\(\displaystyle x=\dfrac{5}{H}\)
\(\displaystyle x=5+H\)
\(\displaystyle H=x+5\)
\(\displaystyle H=5x\)
\(\displaystyle H=\dfrac{5}{x}\)
\(\displaystyle x-H=5\)
\(\displaystyle H-x=5\)
\(\displaystyle \dfrac{x}{H}=5\)
\(\displaystyle \dfrac{H}{x}=5\)
\(\displaystyle \dfrac{\log x}{\log H} = 5\)
\(\displaystyle \log x - \log H =\log 5\)
\(\displaystyle \log x + \log 5 = \log H\)
CheckpointA.10.17.
From the list above, match all the correct algebraic expressions to the phrase "\(x\) is 5 times as large as \(H\text{.}\)"
Answer.
(a), (i), (l)
CheckpointA.10.18.
From the list above, match all the correct algebraic expressions to the phrase "\(x\) is 5 more than \(H\text{.}\)"
Answer.
(c), (g)
SubsectionA.10.3The Natural Base
SubsubsectionA.10.3.1Graphs of \(y=e^x\) and \(y=\ln {(x)}\)
The graphs of the natural exponential function and the natural log function have some special properties.
CheckpointA.10.19.
Use technology to graph \(~f(x)=e^x~\) and \(~y=x+1~\) in a window with \(~-2 \le {(x)} \le 3~\) and \(~-1 \le y \le 4~\text{.}\) What do you notice about the two graphs?
Answer.
The line is tangent to the graph at \((0,1)\text{.}\)
CheckpointA.10.20.
Use technology to graph \(~f(x)=\ln x~\) and \(~y=x-1~\) in a window with \(~-1 \le {(x)} \le 4~\) and \(~-2 \le y \le 3~\text{.}\) What do you notice about the two graphs?
Answer.
The line is tangent to the graph at \((1,0)\text{.}\)
SubsubsectionA.10.3.2Using growth and decay laws with base \(e\)
We can write exponential growth and decay laws using base \(e\text{.}\)
Using base \(e\text{,}\) we write \(~P(t) = P_0 e^{kt},~\) where \(e^k = 1.08.\) (You can see this by evaluating each growth law at \(t=1\text{.}\)) So we solve for \(k\text{.}\)
\begin{align*}
e^k \amp = 1.08 \amp \amp \blert{\text{Take the natural log of both sides.}}\\
\ln (e^k) \amp = \ln (1.08) \amp \amp \blert{\text{Simplify both sides.}}\\
k \amp = 0.0770
\end{align*}
The growth law is \(~P(t) = P_0 e^{0.077t}\text{.}\)
ExampleA.10.22.
A radioactive isotope decays according to the formula \(~N(t)=N_0 e^{-0.016t},~\) where \(t\) is in hours. Find its percent rate of decay.
Solution.
First we write the decay law in the form \(~N(t)=N_0 b^t,~\) where \(~b=e^k.~\)
In this case, \(~k=-0.016,~\) so \(~b=e^{-0.016} = 0.9841.~\) Now, \(~b=1-r,~\) and solving for \(r\) we find \(~r=-0.0159.~\) The rate of decay is approximately 16% per hour.
CheckpointA.10.23.
A virus spreads in the population at a rate of 19.5% daily. Write its growth law using base \(e\text{.}\)
Answer.
\(P(t) = P_0 e^{0.178t}\)
CheckpointA.10.24.
Sea ice is decreasing at a rate of 12.85% per decade. Write its decay law using base \(e\text{.}\)
Answer.
\(Q(t) = Q_0 e^{-0.1375t}\)
CheckpointA.10.25.
In 2020, the world population was growing according to the formula \(~P(t)=P_0 e^{0.0488t},~\) where \(t\) is in years. Find its percent rate of growth.
Answer.
5%
CheckpointA.10.26.
Since 1984, the population of cod has decreased annually according to the formula \(~N(t)=N_0 e^{-0.1863t}.~\) Find its percent rate of decay.
