Intermediate Algebra 2e · free preview

10.3 Evaluate and Graph Logarithmic Functions

Learning objectives: Convert between exponential and logarithmic form; Evaluate logarithmic functions; Graph Logarithmic functions; Solve logarithmic equations; Use logarithmic models in applications

1. Learning objectives

Learning objectives

Learning objectives

  • Convert between exponential and logarithmic form
  • Evaluate logarithmic functions
  • Graph Logarithmic functions
  • Solve logarithmic equations
  • Use logarithmic models in applications

We have spent some time finding the inverse of many functions. It works well to ‘undo’ an operation with another operation. Subtracting ‘undoes’ addition, multiplication ‘undoes’ division, taking the square root ‘undoes’ squaring.

As we studied the exponential function, we saw that it is one-to-one as its graphs pass the horizontal line test. This means an exponential function does have an inverse. If we try our algebraic method for finding an inverse, we run into a problem.

Rewritewith y=f(x).Interchangethevariables x and y.f(x)=axy=axx=aySolvefor y.Oops!Wehavenowaytosolvefor y!\begin{array}{lccllcclcc} \begin{array}{l} \mathrm{Rewrite with}\,y=f(x). \\ \mathrm{Interchange the variables}\,x\,\mathrm{and}\,y. \\ \end{array} & & & \begin{array}{rll} f(x) & = & {a}^{x} \\ y & = & {a}^{x} \\ x & = & {a}^{y} \\ \end{array} \\ \mathrm{Solve for}\,y. & & & \mathrm{Oops! We have no way to solve for}\,y! \\ \end{array}

To deal with this we define the logarithm function with base a to be the inverse of the exponential function f(x)=ax.f(x)={a}^{x}. We use the notation f−1(x)=logax{f}^{-1}(x)={\mathrm{log}}_{a}x and say the inverse function of the exponential function is the logarithmic function.

The function f(x)=logaxf(x)={\mathrm{log}}_{a}x is the logarithmic function with base aa, where a>0,a>0,x>0,x>0, and a≠1.a\ne{}1.

y=logax isequivalentto x=ayy={\mathrm{log}}_{a}x\,\mathrm{is equivalent to}\,x={a}^{y}

2. Convert Between Exponential and Logarithmic Form

Convert Between Exponential and Logarithmic Form

Since the equations y=logaxy={\mathrm{log}}_{a}x and x=ayx={a}^{y} are equivalent, we can go back and forth between them. This will often be the method to solve some exponential and logarithmic equations. To help with converting back and forth let’s take a close look at the equations. See Figure 10.3. Notice the positions of the exponent and base.

If we realize the logarithm is the exponent it makes the conversion easier. You may want to repeat, “base to the exponent give us the number.”

3. Example 10.18

Example 10.18

Example 10.18

Convert to logarithmic form: ⓐ 23=8,{2}^{3}=8, ⓑ 512=5,{5}^{\frac{1}{2}}=\sqrt{5}, and ⓒ (12)4=116.{(\frac{1}{2})}^{4}=\frac{1}{16}.

In the next example we do the reverse—convert logarithmic form to exponential form.

4. Example 10.19

Example 10.19

Example 10.19

Convert to exponential form: ⓐ 2=log864,2={\mathrm{log}}_{8}64, ⓑ 0=log41,0={\mathrm{log}}_{4}1, and ⓒ −3=log1011000.-3={\mathrm{log}}_{10}\frac{1}{1000}.

5. Evaluate Logarithmic Functions

Evaluate Logarithmic Functions

We can solve and evaluate logarithmic equations by using the technique of converting the equation to its equivalent exponential equation.

6. Example 10.20

Example 10.20

Example 10.20

Find the value of x: ⓐ logx36=2,{\mathrm{log}}_{x}36=2, ⓑ log4x=3,{\mathrm{log}}_{4}x=3, and ⓒ log1218=x.{\mathrm{log}}_{\frac{1}{2}}\frac{1}{8}=x.

Solution

ⓐ

 logx36 = 2\,{\mathrm{log}}_{x}36\,=\,2

Convert to exponential form. ·  x2 = 36\,{x}^{2}\,=\,36 Solve the quadratic. ·  x=6, x=−6\,x=6,\,x=-6 The base of a logarithmic function must be positive, so we eliminate x=−6x=-6. ·  x = 6 Therefore, log636=2.\,x\,=\,6\,\mathrm{Therefore,}\,{\mathrm{log}}_{6}36=2.

ⓑ

 log4x = 3\,{\mathrm{log}}_{4}x\,=\,3

Convert to exponential form. ·  43 = x\,{4}^{3}\,=\,x Simplify. ·  x = 64 Therefore, log464 = 3.\,x\,=\,64\,\mathrm{Therefore,}\,{\mathrm{log}}_{4}64\,=\,3.

ⓒ

log1218 = x{\mathrm{log}}_{\frac{1}{2}}\frac{1}{8}\,=\,x

Convert to exponential form. ·  (12)x = 18\,{(\frac{1}{2})}^{x}\,=\,\frac{1}{8} Rewrite 18\frac{1}{8} as (12)3{(\frac{1}{2})}^{3}. ·  (12)x = (12)3\,{(\frac{1}{2})}^{x}\,=\,{(\frac{1}{2})}^{3} With the same base, the exponents must be equal. ·  x = 3 Therefore, log1218=3\,x\,=\,3\,\mathrm{Therefore,}\,{\mathrm{log}}_{\frac{1}{2}}\frac{1}{8}=3

When see an expression such as log327,{\mathrm{log}}_{3}27, we can find its exact value two ways. By inspection we realize it means "3\mathrm{"}3 to what power will be 27"?27\mathrm{"}? Since 33=27,{3}^{3}=27, we know log327=3.{\mathrm{log}}_{3}27=3. An alternate way is to set the expression equal to xx and then convert it into an exponential equation.

7. Example 10.21

Example 10.21

Example 10.21

Find the exact value of each logarithm without using a calculator: ⓐ log525,{\mathrm{log}}_{5}25, ⓑ log93,{\mathrm{log}}_{9}3, and ⓒ log2116.{\mathrm{log}}_{2}\frac{1}{16}.

Solution

ⓐ

log525{\mathrm{log}}_{5}25

5 to what power will be 25? · log525 = 2{\mathrm{log}}_{5}25\,=\,2

Or

Set the expression equal to xx. · log525 = x{\mathrm{log}}_{5}25\,=\,x Change to exponential form. ·  5x = 25\,{5}^{x}\,=\,25 Rewrite 25 as 52{5}^{2}. ·  5x = 52\,{5}^{x}\,=\,{5}^{2} With the same base the exponents must be equal. ·  x = 2 Therefore, log525=2.\,x\,=\,2\,\mathrm{Therefore,}\,{\mathrm{log}}_{5}25=2.

ⓑ

log93{\mathrm{log}}_{9}3

Set the expression equal to xx. · log93 = x{\mathrm{log}}_{9}3\,=\,x Change to exponential form. ·  9x = 3\,{9}^{x}\,=\,3 Rewrite 9 as 32{3}^{2}. · (32)x = 31{({3}^{2})}^{x}\,=\,{3}^{1} Simplify the exponents. ·  32x = 31\,{3}^{2x}\,=\,{3}^{1} With the same base the exponents must be equal. ·  2x = 1\,2x\,=\,1 Solve the equation. ·  x = 12 Therefore, log93=12.\,x\,=\,\frac{1}{2}\,\mathrm{Therefore,}\,{\mathrm{log}}_{9}3=\frac{1}{2}.

ⓒ

log2116{\mathrm{log}}_{2}\frac{1}{16}

Set the expression equal to xx. · log2116 = x{\mathrm{log}}_{2}\frac{1}{16}\,=\,x Change to exponential form. ·  2x = 116\,{2}^{x}\,=\,\frac{1}{16} Rewrite 16 as 24{2}^{4}. ·  2x = 124\,{2}^{x}\,=\,\frac{1}{{2}^{4}}

 2x = 2−4\,{2}^{x}\,=\,{2}^{-4}

With the same base the exponents must be equal. ·  x = −4 Therefore, log2116=−4.\,x\,=\,-4\,\mathrm{Therefore,}\,{\mathrm{log}}_{2}\frac{1}{16}=-4.

8. Graph Logarithmic Functions

Graph Logarithmic Functions

To graph a logarithmic function y=logax,y={\mathrm{log}}_{a}x, it is easiest to convert the equation to its exponential form, x=ay.x={a}^{y}. Generally, when we look for ordered pairs for the graph of a function, we usually choose an x-value and then determine its corresponding y-value. In this case you may find it easier to choose y-values and then determine its corresponding x-value.

9. Example 10.22

Example 10.22

Example 10.22

Graph y=log2x.y={\mathrm{log}}_{2}x.

Solution

To graph the function, we will first rewrite the logarithmic equation, y=log2x,y={\mathrm{log}}_{2}x, in exponential form, 2y=x.{2}^{y}=x.

We will use point plotting to graph the function. It will be easier to start with values of y and then get x.

yy · 2y=x{2}^{y}=x · (x,y)(x,y)−2-2 · 2−2=122=14{2}^{-2}=\frac{1}{{2}^{2}}=\frac{1}{4} · (14,−2)(\frac{1}{4},-2)−1-1 · 2−1=121=12{2}^{-1}=\frac{1}{{2}^{1}}=\frac{1}{2} · (12,−1)(\frac{1}{2},-1) 0 · 20=1{2}^{0}=1 · (1,0)(1,0) 1 · 21=2{2}^{1}=2 · (2,1)(2,1) 2 · 22=4{2}^{2}=4 · (4,2)(4,2) 3 · 23=8{2}^{3}=8 · (8,3)(8,3)

The graphs of y=log2x,y={\mathrm{log}}_{2}x,y=log3x,y={\mathrm{log}}_{3}x, and y=log5xy={\mathrm{log}}_{5}x are the shape we expect from a logarithmic function where a>1.a>1.

We notice that for each function the graph contains the point (1,0).(1,0). This make sense because 0=loga10={\mathrm{log}}_{a}1 means a0=1{a}^{0}=1 which is true for any a.

The graph of each function, also contains the point (a,1).(a,1). This makes sense as 1=logaa1={\mathrm{log}}_{a}a means a1=a.{a}^{1}=a. which is true for any a.

Notice too, the graph of each function y=logaxy={\mathrm{log}}_{a}x also contains the point (1a,−1).(\frac{1}{a},-1). This makes sense as −1=loga1a-1={\mathrm{log}}_{a}\frac{1}{a} means a−1=1a,{a}^{-1}=\frac{1}{a}, which is true for any a.

Look at each graph again. Now we will see that many characteristics of the logarithm function are simply ’mirror images’ of the characteristics of the corresponding exponential function.

What is the domain of the function? The graph never hits the y-axis. The domain is all positive numbers. We write the domain in interval notation as (0,∞).(0,\infty).

What is the range for each function? From the graphs we can see that the range is the set of all real numbers. There is no restriction on the range. We write the range in interval notation as (−∞,∞).(-\infty,\infty).

When the graph approaches the y-axis so very closely but will never cross it, we call the line x=0,x=0, the y-axis, a vertical asymptote.

Domain · (0,∞)(0,\infty) Range · (−∞,∞)(-\infty,\infty) x−interceptx-\mathrm{intercept} · (1,0)(1,0) y−intercepty-\mathrm{intercept} · None Contains · (a,1),(a,1),(1a,−1)(\frac{1}{a},-1) Asymptote · y−axisy-\mathrm{axis}

Our next example looks at the graph of y=logaxy={\mathrm{log}}_{a}x when 0<a<1.0<a<1.

10. Example 10.23

Example 10.23

Example 10.23

Graph y=log13x.y={\mathrm{log}}_{\frac{1}{3}}x.

Solution

To graph the function, we will first rewrite the logarithmic equation, y=log13x,y={\mathrm{log}}_{\frac{1}{3}}x, in exponential form, (13)y=x.{(\frac{1}{3})}^{y}=x.

We will use point plotting to graph the function. It will be easier to start with values of y and then get x.

yy · (13)y=x{(\frac{1}{3})}^{y}=x · (x,y)(x,y)−2-2 · (13)−2=32=9{(\frac{1}{3})}^{-2}={3}^{2}=9 · (9,−2)(9,-2)−1-1 · (13)−1=31=3{(\frac{1}{3})}^{-1}={3}^{1}=3 · (3,−1)(3,-1) 0 · (13)0=1{(\frac{1}{3})}^{0}=1 · (1,0)(1,0) 1 · (13)1=13{(\frac{1}{3})}^{1}=\frac{1}{3} · (13,1)(\frac{1}{3},1) 2 · (13)2=19{(\frac{1}{3})}^{2}=\frac{1}{9} · (19,2)(\frac{1}{9},2) 3 · (13)3=127{(\frac{1}{3})}^{3}=\frac{1}{27} · (127,3)(\frac{1}{27},3)

Now, let’s look at the graphs y=log12x, y=log13xy={\mathrm{log}}_{\frac{1}{2}}x,\,y={\mathrm{log}}_{\frac{1}{3}}x and y=log14xy={\mathrm{log}}_{\frac{1}{4}}x, so we can identify some of the properties of logarithmic functions where 0<a<1.0<a<1.

The graphs of all have the same basic shape. While this is the shape we expect from a logarithmic function where 0<a<1.0<a<1.

We notice, that for each function again, the graph contains the points,(1,0),(1,0),(a,1),(a,1),(1a,−1).(\frac{1}{a},-1). This make sense for the same reasons we argued above.

We notice the domain and range are also the same—the domain is (0,∞)(0,\infty) and the range is (−∞,∞).(-\infty,\infty). The yy-axis is again the vertical asymptote.

We will summarize these properties in the chart below. Which also include when a>1.a>1.

when a>1a>1 · when 0<a<10<a<1 Domain · (0,∞)(0,\infty) · Domain · (0,∞)(0,\infty) Range · (−∞,∞)(-\infty,\infty) · Range · (−∞,∞)(-\infty,\infty) xx-intercept · (1,0)(1,0) · xx-intercept · (1,0)(1,0) yy-intercept · none · yy-intercept · None Contains · (a,1),(a,1),(1a,−1)(\frac{1}{a},-1) · Contains · (a,1),(a,1),(1a,−1)(\frac{1}{a},-1) Asymptote · yy-axis · Asymptote · yy-axis Basic shape · increasing · Basic shape · Decreasing

We talked earlier about how the logarithmic function f−1(x)=logax{f}^{-1}(x)={\mathrm{log}}_{a}x is the inverse of the exponential function f(x)=ax.f(x)={a}^{x}. The graphs in Figure 10.4 show both the exponential (blue) and logarithmic (red) functions on the same graph for both a>1a>1 and 0<a<1.0<a<1.

Notice how the graphs are reflections of each other through the line y=x.y=x. We know this is true of inverse functions. Keeping a visual in your mind of these graphs will help you remember the domain and range of each function. Notice the x-axis is the horizontal asymptote for the exponential functions and the y-axis is the vertical asymptote for the logarithmic functions.

11. Solve Logarithmic Equations

Solve Logarithmic Equations

When we talked about exponential functions, we introduced the number e. Just as e was a base for an exponential function, it can be used a base for logarithmic functions too. The logarithmic function with base e is called the natural logarithmic function. The function f(x)=logexf(x)={\mathrm{log}}_{e}x is generally written f(x)=ln xf(x)=\mathrm{ln}\,x and we read it as “el en of x."x.\mathrm{"}

The function f(x)=ln xf(x)=\mathrm{ln}\,x is the natural logarithmic function with base e,e, where x>0.x>0.

y=ln x isequivalentto x=eyy=\mathrm{ln}\,x\,\mathrm{is equivalent to}\,x={e}^{y}

When the base of the logarithm function is 10, we call it the common logarithmic function and the base is not shown. If the base a of a logarithm is not shown, we assume it is 10.

The function f(x)=log xf(x)=\mathrm{log}\,x is the common logarithmic function with base1010, where x>0.x>0.

y=log x isequivalentto x=10yy=\mathrm{log}\,x\,\mathrm{is equivalent to}\,x={10}^{y}

To solve logarithmic equations, one strategy is to change the equation to exponential form and then solve the exponential equation as we did before. As we solve logarithmic equations, y=logaxy={\mathrm{log}}_{a}x, we need to remember that for the base a, a>0a>0 and a≠1.a\ne{}1. Also, the domain is x>0.x>0. Just as with radical equations, we must check our solutions to eliminate any extraneous solutions.

12. Example 10.24

Example 10.24

Example 10.24

Solve: ⓐ loga49=2{\mathrm{log}}_{a}49=2 and ⓑ ln x=3.\mathrm{ln}\,x=3.

Solution

ⓐ

loga49 = 2{\mathrm{log}}_{a}49\,=\,2

Rewrite in exponential form. ·  a2 = 49\,{a}^{2}\,=\,49 Solve the equation using the square root property. ·  a = ±7\,a\,=\,\pm{}7 The base cannot be negative, so we eliminate a=−7.a=-7. · a=7, a=−7a=7,\,a=-7

Check.

a=7loga49=2log749=?272=?4949=49✓\begin{array}{crll} a=7 & {\mathrm{log}}_{a}49 & = & 2 \\ & {\mathrm{log}}_{7}49 & \overset{?}{=} & 2 \\ & {7}^{2} & \overset{?}{=} & 49 \\ & 49 & = & 49\mathrm{✓} \\ \end{array}

ⓑ

 ln x = 3\,\mathrm{ln}\,x\,=\,3

Rewrite in exponential form. ·  e3 = x\,{e}^{3}\,=\,x

Check.

x=e3ln x=3ln e3=?3e3=e3✓\begin{array}{crll} x={e}^{3} & \mathrm{ln}\,x & = & 3 \\ & \mathrm{ln}\,{e}^{3} & \overset{?}{=} & 3 \\ & {e}^{3} & = & {e}^{3}\mathrm{✓} \\ \end{array}

13. Example 10.25

Example 10.25

Example 10.25

Solve: ⓐ log2(3x−5)=4{\mathrm{log}}_{2}(3x-5)=4 and ⓑ ln e2x=4.\mathrm{ln}\,{e}^{2x}=4.

Solution

ⓐ

log2(3x−5) = 4 {\mathrm{log}}_{2}(3x-5)\,=\,4\,

Rewrite in exponential form. · 24 = 3x−5{2}^{4}\,=\,3x-5 Simplify. · 16 = 3x−516\,=\,3x-5 Solve the equation. · 21 = 3x 21\,=\,3x\,

7 = x 7\,=\,x\,

Check.

x=7log2(3x−5)=4log2(3⋅7−5)=?4log2(16)=?424=?1616=16✓ \begin{array}{lrll} x=7 & {\mathrm{log}}_{2}(3x-5) & = & 4 \\ & {\mathrm{log}}_{2}(3\cdot{}7-5) & \overset{?}{=} & 4 \\ & {\mathrm{log}}_{2}(16) & \overset{?}{=} & 4 \\ & {2}^{4} & \overset{?}{=} & 16 \\ & 16 & = & 16\mathrm{✓}\, \\ \end{array}

ⓑ

ln e2x = 4 \mathrm{ln}\,{e}^{2x}\,=\,4\,

Rewrite in exponential form. · e4 = e2x{e}^{4}\,=\,{e}^{2x} Since the bases are the same the exponents are equal. · 4 = 2x 4\,=\,2x\, Solve the equation. · 2 = x 2\,=\,x\,

Check.

x=2ln e2x=4ln e2⋅2=?4ln e4=?4e4=e4✓\begin{array}{lrll} x=2 & \mathrm{ln}\,{e}^{2x} & = & 4 \\ & \mathrm{ln}\,{e}^{2\cdot{}2} & \overset{?}{=} & 4 \\ & \mathrm{ln}\,{e}^{4} & \overset{?}{=} & 4 \\ & {e}^{4} & = & {e}^{4}\mathrm{✓} \\ \end{array}

14. Use Logarithmic Models in Applications

Use Logarithmic Models in Applications

There are many applications that are modeled by logarithmic equations. We will first look at the logarithmic equation that gives the decibel (dB) level of sound. Decibels range from 0, which is barely audible to 160, which can rupture an eardrum. The 10−12{10}^{-12} in the formula represents the intensity of sound that is barely audible.

The loudness level, D, measured in decibels, of a sound of intensity, I, measured in watts per square inch is

D=10 log(I10−12)D=10\,\mathrm{log}(\frac{I}{{10}^{-12}})

15. Example 10.26

Example 10.26

Example 10.26

Extended exposure to noise that measures 85 dB can cause permanent damage to the inner ear which will result in hearing loss. What is the decibel level of music coming through ear phones with intensity 10−2{10}^{-2} watts per square inch?

Solution

The magnitude RR of an earthquake is measured by a logarithmic scale called the Richter scale. The model is R=log I,R=\mathrm{log}\,I, where II is the intensity of the shock wave. This model provides a way to measure earthquake intensity.

The magnitude R of an earthquake is measured by R=log I,R=\mathrm{log}\,I, where I is the intensity of its shock wave.

16. Example 10.27

Example 10.27

Example 10.27

In 1906, San Francisco experienced an intense earthquake with a magnitude of 7.8 on the Richter scale. Over 80% of the city was destroyed by the resulting fires. In 2014, Los Angeles experienced a moderate earthquake that measured 5.1 on the Richter scale and caused $108 million dollars of damage. Compare the intensities of the two earthquakes.

Solution

To compare the intensities, we first need to convert the magnitudes to intensities using the log formula. Then we will set up a ratio to compare the intensities.

Convert the magnitudes to intensities. ·  R=log I\,R=\mathrm{log}\,I 1906earthquake\,\mathrm{1906 earthquake} · 7.8=log I7.8=\mathrm{log}\,I Converttoexponentialform.\,\mathrm{Convert to exponential form.} ·  I=107.8\,I={10}^{7.8} 2014earthquake\,\mathrm{2014 earthquake} · 5.1=log I5.1=\mathrm{log}\,I Converttoexponentialform.\,\mathrm{Convert to exponential form.} ·  I=105.1\,I={10}^{5.1} Form a ratio of the intensities. · Intensity for 1906Intensity for 2014\frac{\mathrm{Intensity}\,\mathrm{for}\,1906}{\mathrm{Intensity}\,\mathrm{for}\,2014} Substitute in the values. ·  107.8105.1\,\frac{{10}^{7.8}}{{10}^{5.1}} Divide by subtracting the exponents. ·  102.7\,{10}^{2.7} Evaluate. ·  501\,501

The intensity of the 1906 earthquake was about 501 times the intensity of the 2014 earthquake.

17. Section 10.3 Exercises

Section 10.3 Exercises

Section 10.3 Exercises

Convert Between Exponential and Logarithmic Form

In the following exercises, convert from exponential to logarithmic form.

In the following exercises, convert each logarithmic equation to exponential form.

Evaluate Logarithmic Functions

In the following exercises, find the value of xx in each logarithmic equation.

In the following exercises, find the exact value of each logarithm without using a calculator.

Graph Logarithmic Functions

In the following exercises, graph each logarithmic function.

Solve Logarithmic Equations

In the following exercises, solve each logarithmic equation.

Use Logarithmic Models in Applications

In the following exercises, use a logarithmic model to solve.

Writing Exercises

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

18. Practice 1

  1. Convert to logarithmic form: ⓐ 32=9{3}^{2}=9 ⓑ 712=7{7}^{\frac{1}{2}}=\sqrt{7} ⓒ (13)x=127{(\frac{1}{3})}^{x}=\frac{1}{27}
  1. Convert to logarithmic form: ⓐ 43=64{4}^{3}=64 ⓑ 413=43{4}^{\frac{1}{3}}=\sqrt[3]{4} ⓒ (12)x=132{(\frac{1}{2})}^{x}=\frac{1}{32}
  1. Convert to exponential form: ⓐ 3=log4643={\mathrm{log}}_{4}64 ⓑ 0=logx10={\mathrm{log}}_{x}1 ⓒ −2=log101100-2={\mathrm{log}}_{10}\frac{1}{100}
  1. Convert to exponential form: ⓐ 3=log3273={\mathrm{log}}_{3}27 ⓑ 0=log310={\mathrm{log}}_{3}1 ⓒ −1=log10110-1={\mathrm{log}}_{10}\frac{1}{10}

19. Practice 2

  1. Find the value of x:x: ⓐ logx64=2{\mathrm{log}}_{x}64=2 ⓑ log5x=3{\mathrm{log}}_{5}x=3 ⓒ log1214=x{\mathrm{log}}_{\frac{1}{2}}\frac{1}{4}=x
  1. Find the value of x:x: ⓐ logx81=2{\mathrm{log}}_{x}81=2 ⓑ log3x=5{\mathrm{log}}_{3}x=5 ⓒ log13127=x{\mathrm{log}}_{\frac{1}{3}}\frac{1}{27}=x
  1. Find the exact value of each logarithm without using a calculator: ⓐ log12144{\mathrm{log}}_{12}144 ⓑ log42{\mathrm{log}}_{4}2 ⓒ log2132{\mathrm{log}}_{2}\frac{1}{32}
  1. Find the exact value of each logarithm without using a calculator: ⓐ log981{\mathrm{log}}_{9}81 ⓑ log82{\mathrm{log}}_{8}2 ⓒ log319{\mathrm{log}}_{3}\frac{1}{9}

20. Practice 3

  1. Solve: ⓐ loga121=2{\mathrm{log}}_{a}121=2 ⓑ ln x=7\mathrm{ln}\,x=7
  1. Solve: ⓐ loga64=3{\mathrm{log}}_{a}64=3 ⓑ ln x=9\mathrm{ln}\,x=9
  1. Solve: ⓐ log2(5x−1)=6{\mathrm{log}}_{2}(5x-1)=6 ⓑ ln e3x=6\mathrm{ln}\,{e}^{3x}=6
  1. Solve: ⓐ log3(4x+3)=3{\mathrm{log}}_{3}(4x+3)=3 ⓑ ln e4x=4\mathrm{ln}\,{e}^{4x}=4

21. Practice 4

  1. What is the decibel level of one of the new quiet dishwashers with intensity 10−7{10}^{-7} watts per square inch?
  1. What is the decibel level heavy city traffic with intensity 10−3{10}^{-3} watts per square inch?
  1. In 1906, San Francisco experienced an intense earthquake with a magnitude of 7.8 on the Richter scale. In 1989, the Loma Prieta earthquake also affected the San Francisco area, and measured 6.9 on the Richter scale. Compare the intensities of the two earthquakes.
  1. In 2014, Chile experienced an intense earthquake with a magnitude of 8.2 on the Richter scale. In 2014, Los Angeles also experienced an earthquake which measured 5.1 on the Richter scale. Compare the intensities of the two earthquakes.

22. Practice 5

  1. 25=32{2}^{5}=32
  1. 53=125{5}^{3}=125
  1. 10−2=1100{10}^{-2}=\frac{1}{100}
  1. x13=63{x}^{\frac{1}{3}}=\sqrt[3]{6}

23. Practice 6

  1. 17x=175{17}^{x}=\sqrt[5]{17}
  1. (13)4=181{(\frac{1}{3})}^{4}=\frac{1}{81}
  1. 4−3=164{4}^{-3}=\frac{1}{64}
  1. e3=x{e}^{3}=x

24. Practice 7

  1. 6=log2646={\mathrm{log}}_{2}64
  1. 5=logx325={\mathrm{log}}_{x}32
  1. 0=log710={\mathrm{log}}_{7}1
  1. 1=log991={\mathrm{log}}_{9}9

25. Practice 8

  1. 3=log101,0003={\mathrm{log}}_{10}1,000
  1. x=loge43x={\mathrm{log}}_{e}43
  1. logx121=2{\mathrm{log}}_{x}121=2
  1. logx64=3{\mathrm{log}}_{x}64=3

26. Practice 9

  1. log5x=3{\mathrm{log}}_{5}x=3
  1. log3x=−5{\mathrm{log}}_{3}x=-5
  1. log1319=x{\mathrm{log}}_{\frac{1}{3}}\frac{1}{9}=x
  1. log1981=x{\mathrm{log}}_{\frac{1}{9}}81=x

27. Practice 10

  1. log273{\mathrm{log}}_{27}3
  1. log124{\mathrm{log}}_{\frac{1}{2}}4
  1. log3127{\mathrm{log}}_{3}\frac{1}{27}
  1. log9181{\mathrm{log}}_{9}\frac{1}{81}

28. Practice 11

  1. y=log4xy={\mathrm{log}}_{4}x
  1. y=log7xy={\mathrm{log}}_{7}x
  1. y=log2.5xy={\mathrm{log}}_{2.5}x
  1. y=log15xy={\mathrm{log}}_{\frac{1}{5}}x

29. Practice 12

  1. y=log0.6xy={\mathrm{log}}_{0.6}x
  1. loga81=2{\mathrm{log}}_{a}81=2
  1. loga27=3{\mathrm{log}}_{a}27=3
  1. loga24=3{\mathrm{log}}_{a}24=3

30. Practice 13

  1. ln x=4\mathrm{ln}\,x=4
  1. log2(6x+2)=5{\mathrm{log}}_{2}(6x+2)=5
  1. log3(5x−4)=4{\mathrm{log}}_{3}(5x-4)=4
  1. log4(3x−2)=2{\mathrm{log}}_{4}(3x-2)=2

31. Practice 14

  1. ln e2x=6\mathrm{ln}\,{e}^{2x}=6
  1. log(x2−25)=2\mathrm{log}({x}^{2}-25)=2
  1. log3(x2+2)=3{\mathrm{log}}_{3}({x}^{2}+2)=3
  1. What is the decibel level of a whisper with intensity 10−10{10}^{-10} watts per square inch?

32. Practice 15

  1. What is the decibel level of the sound of a garbage disposal with intensity 10−2{10}^{-2} watts per square inch?
  1. The Los Angeles area experiences many earthquakes. In 1994, the Northridge earthquake measured magnitude of 6.76.7 on the Richter scale. In 2014, Los Angeles also experienced an earthquake which measured 5.15.1 on the Richter scale. Compare the intensities of the two earthquakes.
  1. Explain the difference between common logarithms and natural logarithms.
  1. Explain how to find the log732{\mathrm{log}}_{7}32 on your calculator.

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