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Mathematics LibreTexts

7.4: Properties of the Logarithm

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  • LibreTexts

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Learning Objectives

  • Apply the inverse properties of the logarithm.
  • Expand logarithms using the product, quotient, and power rule for logarithms.
  • 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=logbx if and only if x=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 7.4.1

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 7.4.2

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 logarithm11:

f1(f(x))=logbbx=x and
f(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 7.4.3

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:

Table 7.4.1
logb1=0 logbb=1
log1/bb=1 logb(1b)=1
logbbx=x blogbx=x,x>0

Exercise 7.4.1

Evaluate: log0.00001

Answer

5

www.youtube.com/v/YFsZxKiMCTg

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.

Table 7.4.2
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 logarithms12,

logb(xy)=logbx+logby

and the quotient property of logarithms13,

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 7.4.4

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 7.4.5

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 logarithms14,

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 7.4.6

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,

Table 7.4.3
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)logxlogy

and

log(xy)logxlogy

Example 7.4.7

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 7.4.8:

Expand completely: log310xy2.

Solution

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

log310xy2=log(10xy2)1/3=13log(10xy2)=13(log10+logx+logy2)=13(1+logx+2logy)=13+13logx+23logy

Answer

13+13logx+23logy

Example 7.4.9:

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

Exercise 7.4.2

Expand completely: ln(5y4x)

Answer

ln5+4lny12lnx

www.youtube.com/v/h_Zu-SkATl0

Example 7.4.10:

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

  1. log2(8x2y)
  2. 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 7.4.11:

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 7.4.12:

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)

Exercise 7.4.3

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

Answer

log(25(x+y)3z6)

www.youtube.com/v/o1AVbmS9AU8

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.

Exercise 7.4.4

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
Answer

1. 0

3. 14

5. 10

7. 7

9. 1

11. 1

13. 100

15. 18

17. x2

Exercise 7.4.5

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
Answer

1. e

3. 19

5. 25=32

7. log27

9. 4

Exercise 7.4.6

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(x23yz)
  20. log(2(x+y)3z2)
  21. log(100x3(y+10)3)
  22. log7(x5(y+z)3)
  23. log5(x33yz2)
  24. log(x25y3z2)
Answer

1. log4x+log4y

3. 2+2log3x

5. ln3+2lny

7. log2x2log2y

9. 1+2logx+3logy

11. 3log3xlog3y2log3z

13. 2log5xlog5ylog5z

15. 2+4log6(x+y)

17. log72+12log7x+12log7y

19. 2log3x+13log3ylog3z

21. 2+3logx3log(y+10)

23. 3log5x13log5y23log5z

Exercise 7.4.7

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

  1. log3(27x2y3z)
  2. log3(xy3z)
  3. log3(9x2yz3)
  4. log3(3xyz2)
Answer

1. 3+2a+3b+c

3. 2+2a+b3c

Exercise 7.4.8

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
Answer

1. 2.32

3. 0.71

Exercise 7.4.9

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)
Answer

1. log(3.1)+2525.5

3. ln(6.2)1513.2

Exercise 7.4.10

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
Answer

1. log(xy)

3. log2(5x2y)

5. log2(x3zy2)

7. log[5(x+y)3]

9. ln(xzy6)

11. log(x7yz2)

13. log3(3x2yz)

15. log2(2xy)

17. log2(3xy2)

19. ln((x+y)22z)

21. ln(3xy28z)

23. log2(3yx2z4)

Exercise 7.4.11

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)
Answer

1. log(x21)

3. ln(x+1)

5. log5(x2+2x+4)

7. log(xx5)

Footnotes

11Given b>0 we have logbbx=x and blogbx=x when x>0.

12logb(xy)=logbx+logby; the logarithm of a product is equal to the sum of the logarithm of the factors.

13logb(xy)=logbxlogby; the logarithm of a quotient is equal to the difference of the logarithm of the numerator and the logarithm of the denominator.

14logbxn=nlogbx; the logarithm of a quantity raised to a power is equal to that power times the logarithm of the quantity.]


This page titled 7.4: Properties of the Logarithm is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform.

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