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Sunday, June 30, 2024

21.13 - Properties of Logarithm

In the previous section, we saw some basic details about logarithmic functions. In this section, we will see some of their properties.

Property I: logap=logbplogba
   ♦ On the L.H.S, we have log of p to the base a.
   ♦ On the R.H.S, we have log of p to the base b.
• So this property gives us the method to connect two logs of the same number when the bases are different.
• This property is also known as change of base rule.
• We can write the proof in 5 steps:

1. Let logap=α
Then aα=p
2. Let logbp=β
Then bβ=p
3. Let logba=γ
Then bγ=a
4. Substituting (3) in (1), we get:
(bγ)α=p
bγα=p
5. Substituting (4) in (2), we get:
bβ=p=bγα
β=γα
α=βγ
logap=logbplogba

Property II: logbpq=logbp+logbq
   ♦ On the L.H.S, we have log of a product pq.
   ♦ On the R.H.S, we have addition of two individual logs.
• So this property helps us to simplify logarithmic equations when products are involved.
• We can write the proof in 4 steps:
1. Let logbpq=α
Then bα=pq
2. Let logbp=β
Then bβ=p
3. Let logbq=γ
Then bγ=q
4. Substituting (2) and (3) in (1), we get:
bα=bβ×bγ
bα=bβ+γ
α=β+γ
logbpq=logbp+logbq

Property III: logb(pq)=logbplogbq
   ♦ On the L.H.S, we have log of a ratio pq.
   ♦ On the R.H.S, we have subtraction of one individual log from another.
• So this property helps us to simplify logarithmic equations when ratios are involved.
• We can write the proof in 4 steps:
1. Let logb(pq)=α
Then bα=pq
2. Let logbp=β
Then bβ=p
3. Let logbq=γ
Then bγ=q
4. Substituting (2) and (3) in (1), we get:
bα=bβbγ
bα=bβγ
α=βγ
logbpq=logbplogbq

Property IV: logbp2=2logbp
   ♦ On the L.H.S, we have log of a square.
   ♦ On the R.H.S, we have individual log multiplied by 2.
• So this property helps us to simplify logarithmic equations when squares are involved.
• We can write the proof in 3 steps:
1. Let logb(p×p)=α
Then bα=p×p
2. Let logbp=β
Then bβ=p
3. Substituting (2) in (1), we get:
bα=bβ×bβ
bα=bβ+β
α=β+β
logb(p×p)=logbp+logbp
logbp2=2logbp

Property V: logbp3=3logbp
   ♦ On the L.H.S, we have log of a cube.
   ♦ On the R.H.S, we have individual log multiplied by 3.
• So this property helps us to simplify logarithmic equations when cubes are involved.
• We can write the proof in 3 steps:
1. Let logb(p×p×p)=α
Then bα=p×p×p
2. Let logbp=β
Then bβ=p
3. Substituting (2) in (1), we get:
bα=bβ×bβ×bβ
bα=bβ+β+β
α=β+β+β
logb(p×p×p)=logbp+logbp+logbp
logbp3=3logbp

Property VI: logbpn=nlogbp
   ♦ On the L.H.S, we have log of a nth power.
   ♦ On the R.H.S, we have individual log multiplied by n.
• So this property helps us to simplify logarithmic equations when nth power is involved.
• The proof can be written using the principles of mathematical induction.

Property VII: logpp=1
   ♦ On the L.H.S, we have same number and base.
   ♦ On the R.H.S, we have 1.
We can write the proof as follows:
• Let logpp=α
• Then pα=p
pα=p1
α=1

Property VIII: logbp=1logpb
   ♦ On the L.H.S, we have base b and number p.
   ♦ On the R.H.S, we have base p and number b.
   ♦ So base and number are interchanged.
We can write the proof as follows:
• Applying the change of base rule we get:
logbp=logpplogpb=1logpb 

Property IX: logppx=x
We can write the proof as follows:
• Let logppx=α
• Then pα=px
α=x

• Note that, this property is used when we first begin to learn about logarithms. For example:
log5625=log554=4


Now we will see some solved examples

Solved example 21.43
Express as a single logarithm:
2logx5logy+3logz
Solution:
1. 2 log x = log x2.
2. −5 log y = −log y5.
3. 3 log z = log z3.
4. log x2 − log y5 + log z3 = x2z3y5

Solved example 21.44
Expand using the properties of logarithm:
log53x
Solution:
log53x
= log5(x13) 
= 13log5x

Solved example 21.45
Evaluate log5 28
Solution:
1. We could find the solution easily if it is "25" instead of "28".
2. We cannot directly use the calculator because base is 5.
3. So we will apply the change of base rule. We get:
log528=log1028log105=1.447150.69897=2.07040


In the next section, we will see exponential and logarithmic equations.

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