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7.4 Properties of the Logarithm

Learning Objectives

  1. Apply the inverse properties of the logarithm.
  2. Expand logarithms using the product, quotient, and power rule for logarithms.
  3. Combine logarithms into a single logarithm with coefficient 1.

Logarithms and Their Inverse Properties

Recall the definition of the base-b logarithm: given b>0 where b1,

y=logbxifandonlyifx=by

Use this definition to convert logarithms to exponential form. Doing this, we can derive a few properties:

logb1=0      because    b0=1logbb=1      because    b1=blogb(1b)=1   because    b1=1b

Example 1

Evaluate:

  1. log1
  2. lne
  3. log5(15)

Solution:

  1. When the base is not written, it is assumed to be 10. This is the common logarithm,

    log1=log101=0

  2. The natural logarithm, by definition, has base e,

    lne=logee=1

  3. Because 51=15 we have,

    log5(15)=1

Furthermore, consider fractional bases of the form 1/b where b>1.

log1/bb=1     because    (1b)1=11b1=b1=b

Example 2

Evaluate:

  1. log1/44
  2. log2/3(32)

Solution:

  1. log1/44=1 because (14)1=4
  2. log2/3(32)=1 because (23)1=32

Given an exponential function defined by f(x)=bx, where b>0 and b1, its inverse is the base-b logarithm, f1(x)=logbx. And because f(f1(x))=x and f1(f(x))=x, we have the following inverse properties of the logarithmGiven b>0 we have logbbx=x and blogbx=x when x>0.:

f1(f(x))=logbbx=xandf(f1(x))=blogbx=x,x>0

Since f1(x)=logbx has a domain consisting of positive values (0,), the property blogbx=x is restricted to values where x>0.

Example 3

Evaluate:

  1. log5625
  2. 5log53
  3. eln5

Solution:

Apply the inverse properties of the logarithm.

  1. log5625=log554=4
  2. 5log53=3
  3. eln5=5

In summary, when b>0 and b1, we have the following properties:

logb1=0

logbb=1

log1/bb=1

logb(1b)=1

logbbx=x

blogbx=x, x>0

Try this! Evaluate: log0.00001

Answer: −5

Product, Quotient, and Power Properties of Logarithms

In this section, three very important properties of the logarithm are developed. These properties will allow us to expand our ability to solve many more equations. We begin by assigning u and v to the following logarithms and then write them in exponential form:

logbx=ubu=xlogby=vbv=y

Substitute x=bu and y=bv into the logarithm of a product logb(xy) and the logarithm of a quotient logb(xy). Then simplify using the rules of exponents and the inverse properties of the logarithm.

Logarithm of a Product

Logarithm of a Quotient

logb(xy)=logb(bubv)=logbbu+v=u+v=logbx+logby

logb(xy)=logb(bubv)=logbbuv=uv=logbxlogby

This gives us two essential properties: the product property of logarithmslogb(xy)=logbx+logby; the logarithm of a product is equal to the sum of the logarithm of the factors.,

logb(xy)=logbx+logby

and the quotient property of logarithmslogb(xy)=logbxlogby; the logarithm of a quotient is equal to the difference of the logarithm of the numerator and the logarithm of the denominator.,

logb(xy)=logbxlogby

In words, the logarithm of a product is equal to the sum of the logarithm of the factors. Similarly, the logarithm of a quotient is equal to the difference of the logarithm of the numerator and the logarithm of the denominator.

Example 4

Write as a sum: log2(8x).

Solution:

Apply the product property of logarithms and then simplify.

log2(8x)=log28+log2x=log223+log2x=3+log2x

Answer: 3+log2x

Example 5

Write as a difference: log(x10).

Solution:

Apply the quotient property of logarithms and then simplify.

log(x10)=logxlog10=logx1

Answer: logx1

Next we begin with logbx=u and rewrite it in exponential form. After raising both sides to the nth power, convert back to logarithmic form, and then back substitute.

logbx=ubu=x(bu)n=(x)nlogbxn=nubnu=xnlogbxn=nlogbx

This leads us to the power property of logarithmslogbxn=nlogbx; the logarithm of a quantity raised to a power is equal to that power times the logarithm of the quantity.,

logbxn=nlogbx

In words, the logarithm of a quantity raised to a power is equal to that power times the logarithm of the quantity.

Example 6

Write as a product:

  1. log2x4
  2. log5(x).

Solution:

  1. Apply the power property of logarithms.

    log2x4=4log2x

  2. Recall that a square root can be expressed using rational exponents, x=x1/2. Make this replacement and then apply the power property of logarithms.

    log5(x)=log5x1/2=12log5x

In summary,

Product property of logarithms

logb(xy)=logbx+logby

Quotient property of logarithms

logb(xy)=logbxlogby

Power property of logarithms

logbxn=nlogbx

We can use these properties to expand logarithms involving products, quotients, and powers using sums, differences and coefficients. A logarithmic expression is completely expanded when the properties of the logarithm can no further be applied.

Caution: It is important to point out the following:

log(xy)logxlogyandlog(xy)logxlogy

Example 7

Expand completely: ln(2x3).

Solution:

Recall that the natural logarithm is a logarithm base e, lnx=logex. Therefore, all of the properties of the logarithm apply.

ln(2x3)=ln2+lnx3Productruleforlogarithms=ln2+3lnxPowerruleforlogarithms

Answer: ln2+3lnx

Example 8

Expand completely: log10xy23.

Solution:

Begin by rewriting the cube root using the rational exponent 13 and then apply the properties of the logarithm.

log10xy23=log(10xy2)1/3=13log(10xy2)=13(log10+logx+logy2)=13(1+logx+2logy)=13+13logx+23logy

Answer: 13+13logx+23logy

Example 9

Expand completely: log2((x+1)25y).

Solution:

When applying the product property to the denominator, take care to distribute the negative obtained from applying the quotient property.

log2((x+1)25y)=log2(x+1)2log2(5y)=log2(x+1)2(log25+log2y)Distribute.=log2(x+1)2log25log2y=2log2(x+1)log25log2y

Answer: 2log2(x+1)log25log2y

Caution: There is no rule that allows us to expand the logarithm of a sum or difference. In other words,

log(x±y)logx±logy

Try this! Expand completely: ln(5y4x).

Answer: ln5+4lny12lnx

Example 10

Given that log2x=a, log2y=b, and that log2z=c, write the following in terms of a, b and c:

a. log2(8x2y)

b. log2(2x4z)

Solution:

  1. Begin by expanding using sums and coefficients and then replace a and b with the appropriate logarithm.

    log2(8x2y)=log28+log2x2+log2y=log28+2log2x+log2y=3+2a+b

  2. Expand and then replace a, b, and c where appropriate.

    log2(2x4z)=log2(2x4)log2z1/2=log22+log2x4log2z1/2=log22+4log2x12log2z=1+4a12b

Next we will condense logarithmic expressions. As we will see, it is important to be able to combine an expression involving logarithms into a single logarithm with coefficient 1. This will be one of the first steps when solving logarithmic equations.

Example 11

Write as a single logarithm with coefficient 1:3log3xlog3y+2log35.

Solution:

Begin by rewriting all of the logarithmic terms with coefficient 1. Use the power rule to do this. Then use the product and quotient rules to simplify further.

3log3xlog3y+2log35={log3x3log3y}+log352quotientproperty={log3(x3y)+log325}productproperty=log3(x3y25)=log3(25x3y)

Answer: log3(25x3y)

Example 12

Write as a single logarithm with coefficient 1:12lnx3lnylnz.

Solution:

Begin by writing the coefficients of the logarithms as powers of their argument, after which we will apply the quotient rule twice working from left to right.

12lnx3lnylnz=lnx1/2lny3lnz=ln(x1/2y3)lnz=ln(x1/2y3÷z)=ln(x1/2y31z)=ln(x1/2y3z)or=ln(xy3z)

Answer: ln(xy3z)

Try this! Write as a single logarithm with coefficient 1:3log(x+y)6logz+2log5.

Answer: log(25(x+y)3z6)

Key Takeaways

  • Given any base b>0 and b1, we can say that logb1=0, logbb=1, log1/bb=1 and that logb(1b)=1.
  • The inverse properties of the logarithm are logbbx=x and blogbx=x where x>0.
  • The product property of the logarithm allows us to write a product as a sum: logb(xy)=logbx+logby.
  • The quotient property of the logarithm allows us to write a quotient as a difference: logb(xy)=logbxlogby.
  • The power property of the logarithm allows us to write exponents as coefficients: logbxn=nlogbx.
  • Since the natural logarithm is a base-e logarithm, lnx=logex, all of the properties of the logarithm apply to it.
  • We can use the properties of the logarithm to expand logarithmic expressions using sums, differences, and coefficients. A logarithmic expression is completely expanded when the properties of the logarithm can no further be applied.
  • We can use the properties of the logarithm to combine expressions involving logarithms into a single logarithm with coefficient 1. This is an essential skill to be learned in this chapter.

Topic Exercises

    Part A: Logarithms and Their Inverse Properties

      Evaluate:

    1. log71

    2. log1/22

    3. log1014

    4. log1023

    5. log3310

    6. log66

    7. lne7

    8. ln(1e)
    9. log1/2(12)
    10. log1/55

    11. log3/4(43)

    12. log2/31

    13. 2log2100

    14. 3log31

    15. 10log18

    16. eln23

    17. elnx2

    18. elnex

      Find a:

    1. lna=1

    2. loga=1

    3. log9a=1

    4. log12a=1

    5. log2a=5

    6. loga=13

    7. 2a=7

    8. ea=23

    9. loga45=5

    10. loga10=1

    Part B: Product, Quotient, and Power Properties of Logarithms

      Expand completely.

    1. log4(xy)

    2. log(6x)

    3. log3(9x2)

    4. log2(32x7)

    5. ln(3y2)

    6. log(100x2)

    7. log2(xy2)
    8. log5(25x)
    9. log(10x2y3)

    10. log2(2x4y5)

    11. log3(x3yz2)
    12. log(xy3z2)
    13. log5(1x2yz)
    14. log4(116x2z3)
    15. log6[36(x+y)4]

    16. ln[e4(xy)3]

    17. log7(2xy)

    18. ln(2xy)

    19. log3(x2y3z)
    20. log(2(x+y)3z2)
    21. log(100x3(y+10)3)
    22. log7(x(y+z)35)
    23. log5(x3yz23)
    24. log(x2y3z25)

      Given log3x=a, log3y=b, and log3z=c, write the following logarithms in terms of a, b, and c.

    1. log3(27x2y3z)
    2. log3(xy3z)
    3. log3(9x2yz3)
    4. log3(x3yz2)

      Given logb2=0.43, logb3=0.68, and logb7=1.21, calculate the following. (Hint: Expand using sums, differences, and quotients of the factors 2, 3, and 7.)

    1. logb42

    2. logb(36)

    3. logb(289)

    4. logb21

      Expand using the properties of the logarithm and then approximate using a calculator to the nearest tenth.

    1. log(3.10×1025)

    2. log(1.40×1033)

    3. ln(6.2e15)

    4. ln(1.4e22)

      Write as a single logarithm with coefficient 1.

    1. logx+logy

    2. log3xlog3y

    3. log25+2log2x+log2y

    4. log34+3log3x+12log3y

    5. 3log2x2log2y+12log2z

    6. 4logxlogylog2

    7. log5+3log(x+y)

    8. 4log5(x+5)+log5y

    9. lnx6lny+lnz

    10. log3x2log3y+5log3z

    11. 7logxlogy2logz

    12. 2lnx3lnylnz

    13. 23log3x12(log3y+log3z)

    14. 15(log7x+2log7y)2log7(z+1)

    15. 1+log2x12log2y

    16. 23log3x+13log3y

    17. 13log2x+23log2y

    18. 2log5x+35log5y

    19. ln2+2ln(x+y)lnz

    20. 3ln(xy)lnz+ln5

    21. 13(lnx+2lny)(3ln2+lnz)

    22. 4log2+23logx4log(y+z)

    23. log232log2x+12log2y4log2z

    24. 2log54log5x3log5y+23log5z

      Express as a single logarithm and simplify.

    1. log(x+1)+log(x1)

    2. log2(x+2)+log2(x+1)

    3. ln(x2+2x+1)ln(x+1)

    4. ln(x29)ln(x+3)

    5. log5(x38)log5(x2)

    6. log3(x3+1)log3(x+1)

    7. logx+log(x+5)log(x225)

    8. log(2x+1)+log(x3)log(2x25x3)

Answers

  1. 0

  2. 14

  3. 10

  4. 7

  5. 1

  6. −1

  7. 100

  8. 18

  9. x2

  10. e

  11. 19

  12. 25=32

  13. log27

  14. 4

  1. log4x+log4y

  2. 2+2log3x

  3. ln3+2lny

  4. log2x2log2y

  5. 1+2logx+3logy

  6. 3log3xlog3y2log3z

  7. 2log5xlog5ylog5z

  8. 2+4log6(x+y)

  9. log72+12log7x+12log7y

  10. 2log3x+13log3ylog3z

  11. 2+3logx3log(y+10)

  12. 3log5x13log5y23log5z

  13. 3+2a+3b+c

  14. 2+2a+b3c

  15. 2.32

  16. 0.71

  17. log(3.1)+2525.5

  18. ln(6.2)1513.2

  19. log(xy)

  20. log2(5x2y)

  21. log2(x3zy2)
  22. log[5(x+y)3]

  23. ln(xzy6)
  24. log(x7yz2)
  25. log3(x23yz)
  26. log2(2xy)
  27. log2(xy23)
  28. ln((x+y)22z)
  29. ln(xy238z)
  30. log2(3yx2z4)
  31. log(x21)

  32. ln(x+1)

  33. log5(x2+2x+4)

  34. log(xx5)