7.4: Properties of the Logarithm
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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 b≠1,
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 b−1=1b
Example 7.4.1
Evaluate:
- log1
- lne
- log5(15)
Solution
- When the base is not written, it is assumed to be 10. This is the common logarithm, log1=log101=0
- The natural logarithm, by definition, has base e, lne=logee=1
- Because 5−1=15 we have, log5(15)=−1
Furthermore, consider fractional bases of the form 1/b where b>1.
log1/bb=−1 because (1b)−1=1−1b−1=b1=b
Example 7.4.2
Evaluate:
- log1/44
- log2/3(32)
Solution
- log1/44=−1 because (14)−1=4
- log2/3(32)=−1 because (23)−1=32
Given an exponential function defined by f(x)=bx, where b>0 and b≠1, its inverse is the base-b logarithm, f−1(x)=logbx. And because f(f−1(x))=x and f−1(f(x))=x, we have the following inverse properties of the logarithm11:
f−1(f(x))=logbbx=x and
f(f−1(x))=blogbx=x,x>0
Since f−1(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
- log5625
- 5log53
- eln5
Solution
Apply the inverse properties of the logarithm.
- log5625=log554=4
- 5log53=3
- eln5=5
In summary, when b>0 and b≠1, we have the following properties:
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=u⟹bu=xlogby=v⟹bv=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)=logbbu−v=u−v=logbx−logby |
This gives us two essential properties: the product property of logarithms12,
logb(xy)=logbx+logby
and the quotient property of logarithms13,
logb(xy)=logbx−logby
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)=logx−log10=logx−1
Answer
logx−1
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=u⟹bu=x(bu)n=(x)nlogbxn=nu⟸bnu=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:
- log2x4
- log5(√x)
Solution
- Apply the power property of logarithms.log2x4=4log2x
- 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)=logbx−logby |
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)≠logx⋅logy
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: log3√10xy2.
Solution
Begin by rewriting the cube root using the rational exponent 13 and then apply the properties of the logarithm.
log3√10xy2=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)2−log2(5y)=log2(x+1)2−(log25+log2y)Distribute.=log2(x+1)2−log25−log2y=2log2(x+1)−log25−log2y
Answer
2log2(x+1)−log25−log2y
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(5y4√x)
- Answer
-
ln5+4lny−12lnx
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Example 7.4.10:
Given that log2x=a,log2y=b, and that log2z=c, write the following in terms of a, b, and c:
- log2(8x2y)
- log2(2x4√z)
Solution
- 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
- Expand and then replace a,b, and c where appropriate. log2(2x4√z)=log2(2x4)−log2z1/2=log22+log2x4−log2z1/2=log22+4log2x−12log2z=1+4a−12b
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: 3log3x−log3y+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.
3log3x−log3y+2log35={log3x3−log3y}+log352quotientproperty={log3(x3y)+log325}productproperty=log3(x3y⋅25)=log3(25x3y)
Answer
log3(25x3y)
Example 7.4.12:
Write as a single logarithm with coefficient 1: 12lnx−3lny−lnz.
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.
12lnx−3lny−lnz=lnx1/2−lny3−lnz=ln(x1/2y3)−lnz=ln(x1/2y3÷z)=ln(x1/2y3⋅1z)=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)
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Key Takeaways
- Given any base b>0 and b≠1, 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)=logbx−logby.
- 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:
- log71
- log1/22
- log1014
- log10−23
- log3310
- log66
- lne7
- ln(1e)
- log1/2(12)
- log1/55
- log3/4(43)
- log2/31
- 2log2100
- 3log31
- 10log18
- eln23
- elnx2
- 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:
- lna=1
- loga=−1
- log9a=−1
- log12a=1
- log2a=5
- loga=13
- 2a=7
- ea=23
- loga45=5
- loga10=1
- Answer
-
1. e
3. 19
5. 25=32
7. log27
9. 4
Exercise 7.4.6
Expand completely.
- log4(xy)
- log(6x)
- log3(9x2)
- log2(32x7)
- ln(3y2)
- log(100x2)
- log2(xy2)
- log5(25x)
- log(10x2y3)
- log2(2x4y5)
- log3(x3yz2)
- log(xy3z2)
- log5(1x2yz)
- log4(116x2z3)
- log6[36(x+y)4]
- ln[e4(x−y)3]
- log7(2√xy)
- ln(2x√y)
- log3(x23√yz)
- log(2(x+y)3z2)
- log(100x3(y+10)3)
- log7(x5√(y+z)3)
- log5(x33√yz2)
- log(x25√y3z2)
- Answer
-
1. log4x+log4y
3. 2+2log3x
5. ln3+2lny
7. log2x−2log2y
9. 1+2logx+3logy
11. 3log3x−log3y−2log3z
13. −2log5x−log5y−log5z
15. 2+4log6(x+y)
17. log72+12log7x+12log7y
19. 2log3x+13log3y−log3z
21. 2+3logx−3log(y+10)
23. 3log5x−13log5y−23log5z
Exercise 7.4.7
Given log3x=a,log3y=b, and log3z=c, write the following logarithms in terms of a,b, and and c.
- log3(27x2y3z)
- log3(xy3√z)
- log3(9x2yz3)
- log3(3√xyz2)
- Answer
-
1. 3+2a+3b+c
3. 2+2a+b−3c
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.)
- logb42
- logb(36)
- logb(289)
- logb√21
- 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.
- log(3.10×1025)
- log(1.40×10−33)
- ln(6.2e−15)
- ln(1.4e22)
- Answer
-
1. log(3.1)+25≈25.5
3. ln(6.2)−15≈−13.2
Exercise 7.4.10
Write as a single logarithm with coefficient 1.
- logx+logy
- log3x−log3y
- log25+2log2x+log2y
- log34+3log3x+12log3y
- 3log2x−2log2y+12log2z
- 4logx−logy−log2
- log5+3log(x+y)
- 4log5(x+5)+log5y
- lnx−6lny+lnz
- log3x−2log3y+5log3z
- 7logx−logy−2logz
- 2lnx−3lny−lnz
- 23log3x−12(log3y+log3z)
- 15(log7x+2log7y)−2log7(z+1)
- 1+log2x−12log2y
- 2−3log3x+13log3y
- 13log2x+23log2y
- −2log5x+35log5y
- −ln2+2ln(x+y)−lnz
- −3ln(x−y)−lnz+ln5
- 13(lnx+2lny)−(3ln2+lnz)
- 4log2+23logx−4log(y+z)
- log23−2log2x+12log2y−4log2z
- 2log54−log5x−3log5y+23log5z
- Answer
-
1. log(xy)
3. log2(5x2y)
5. log2(x3√zy2)
7. log[5(x+y)3]
9. ln(xzy6)
11. log(x7yz2)
13. log3(3√x2√yz)
15. log2(2x√y)
17. log2(3√xy2)
19. ln((x+y)22z)
21. ln(3√xy28z)
23. log2(3√yx2z4)
Exercise 7.4.11
Express as a single logarithm and simplify.
- log(x+1)+log(x−1)
- log2(x+2)+log2(x+1)
- ln(x2+2x+1)−ln(x+1)
- ln(x2−9)−ln(x+3)
- log5(x3−8)−log5(x−2)
- log3(x3+1)−log3(x+1)
- logx+log(x+5)−log(x2−25)
- log(2x+1)+log(x−3)−log(2x2−5x−3)
- Answer
-
1. log(x2−1)
3. ln(x+1)
5. log5(x2+2x+4)
7. log(xx−5)
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)=logbx−logby; 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.]