LOGARITHM PROBLEMS

In this section, you will learn how to solve logarithm problems. 

Before look at the problems, if you want to learn about logarithms in detail, 

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Logarithm Problems

Problem 1 :

Find the logarithm of 64 to the base 2√2.

Solution : 

Write 64 as in terms of 2√2. 

64  =  2⁶

64  =  2⁴⁺²

64  =  2⁴ ⋅ 2²

64  =  2⁴ ⋅ [(√2)²]²

64  =  2⁴ ⋅ (√2)⁴

64  =  (2√2)⁴

Then, 

log_2√264  =  log_2√2(2√2)⁴

log_2√264  =  4log_2√2(2√2)

log_2√264  =  4(1)

log_2√264  =  4

Problem 2 :

If log_abc  =  x, log_bca  =  y and log_cab  =  z, then find the value of

1/(x+1) + 1/(y+1) + 1/(z+1)

Solution : 

x + 1  =  log_abc + log_aa  =  log_aabc

y + 1  =  log_bca + log_bc  =  log_babc

z + 1  =  log_cab + log_cc  =  log_cabc

1/(x+1)  =  1 / log_aabc  =  log_abca

1/(y+1)  =  1/log_babc  =  log_abcb

1/(z+1)  =  1/log_cabc  =  log_abcc

1/(x+1) + 1/(y+1) + 1/(z+1)  =  log_abca + log_abcb + log_abcc

1/(x+1) + 1/(y+1) + 1/(z+1)  =  log_abcabc

1/(x+1) + 1/(y+1) + 1/(z+1)  =  1

Problem 3 :

If a = log₂₄12, b = ₂₄36 and c = log₄₈36, then find the value of

1 + abc

Solution : 

1 + abc  =  1 + log₂₄12 ⋅ log₃₆24 ⋅ log₄₈36

1 + abc  =  1 + log₃₆12 ⋅ log₄₈36

1 + abc  =  1 + log₄₈12

1 + abc  =  log₄₈48 + log₄₈12

1 + abc  =  log₄₈(48 ⋅ 12)

1 + abc  =  log₄₈(2 ⋅ 12)²

1 + abc  =  2log₄₈24

1 + abc  =  2log₃₆24 ⋅ log₄₈36

1 + abc  =  2bc

Problem 4 :

Find the value of log 0.0001 to the base 0.1.

Solution : 

log_0.10.0001  =  log_0.1(0.1)⁴

log_0.10.0001  =  4log_0.10.1

log_0.10.0001  =  4(1)

log_0.10.0001  =  4

Problem 5 :

If 2logx  =  4log3,  then find the value of 'x'. 

Solution : 

2logx  =  4log3

Divide each side by 2.

logx  =  (4log3) / 2

logx  =  2log3

logx  =  log3²

logx  =  log9

x  =  9

Problem 6 :

Find the value of log_√264.

Solution : 

log_√264  =  log_√2(2)⁶

log_√264  =  6log_√2(2)

log_√264  =  6log_√2(√2)²

log_√264  =  6 ⋅ 2log_√2(√2)

log_√264  =  12 ⋅ 2(1)

log_√264  =  12

Problem 7 :

Find the value of log (1/81) to the base 9. 

Solution : 

log₉(1/81)  =  log₉1 - log₉81

log₉(1/81)  =  0 - log₉(9)²

log₉(1/81)  =  -2log₉9

log₉(1/81)  =  -2(1)

log₉(1/81)  =  -2

Problem 8 :

Find the value of log(0.0625) to the base 2. 

Solution : 

log₂(0.0625)  =  log₂(0.5)⁴

log₂(0.0625)  =  4log₂(0.5)

log₂(0.0625)  =  4log₂(1/2)

log₂(0.0625)  =  4(log₂1 - log₂2)

log₂(0.0625)  =  4(0 - 1)

log₂(0.0625)  =  4(-1)

log₂(0.0625)  =  -4

Problem 9 :

Given log2 = 0.3010 and log3 = 0.4771, find the value of log6.

Solution : 

log6  =  log(2 ⋅ 3)

log6  =  log2 + log3

Substitute the values of log2 and log3. 

log6  =  0.3010 + 0.4771

log6  =  0.7781

Problem 10 :

Find the value of log(0.3) to the base 9. 

Solution : 

log₉(0.3)  =  log₉(1/3)

log₉(0.3)  =  log₉1 - log₉3

log₉(0.3)  =  0 - log₉3

log₉(0.3)  =   - log₉3

log₉(0.3)  =  - 1 / log₃9

log₉(0.3)  =  - 1 / log₃3²

log₉(0.3)  =  - 1 / 2log₃3

log₉(0.3)  =  - 1 / 2(1)

log₉(0.3)  =  -1/2

After having gone through the stuff given above, we hope that the students would have understood how to solve problems on logarithms. 

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