FINDING LOGARITHMS WITHOUT USING TABLE







FINDING LOGARITHMS WITHOUT USING TABLES.



Example 1:

log₁₀2 = 0.3010 and log₁₀3 = 0.4771. Evaluate log₁₀4.5

Solution:

log₁₀4.5 = 

log₁₀45/10 = 

log₁₀9/2 = log₁₀9 - 

log₁₀2 

= log₁₀3² - 

log₁₀2 = 2 x 0.4771 - 

0.3010 

= 0.6532


Example 2:


if log₁₀2 = 0.30103, find without using logarithms tables the number of digit in 2⁶⁴ 

Solutions:

2⁶⁴ = y

log₂y = 64

logy/log2 = 64
logy/ 0.30103 =64
64 x 0.30103 = logy
10¹⁹·²⁶⁵⁹ = 20





Example 3:

Given that log₁₀5 = 0.6990 and log₁₀7 = 0.8451. Without using tables, find the value of log₁₀140. Hence, solve x⁰·⁹⁶⁴¹ = 140

Solution:


log₁₀140 = log₁₀700/5 = log₁₀700 - log₁₀5 = 
log₁₀ 7 x 100 - log₁₀5=
log₁₀7 + log₁₀10  - log₁₀5 
substitute the logs into the equation 
= 0.8451 + 2 - 0.6990 = 2.1461

Hence,     x⁰·⁹⁶⁴¹ = log₁₀140 ⇒ take log of both side
log₁₀x⁰·⁹⁶⁴¹ = log₁₀140

0.9641log₁₀x = 2.1461
log₁₀x = 2.1461
              ------------
              0.9641
          = 2.1461
x = 10²·²²⁶⁰ 
x = antilog of 2.2260
x = 168.27



Example 4:

Given that log₁₀2 = 0.30103, and log₁₀3 = 0.47712

Evaluate log₁₀15

Solution :

log₁₀15 = log₁₀30/2 = log₁₀30 - log₁₀2
= log₁₀3 x 10 - log₁₀2
= log₁₀3 + log₁₀10 - log₁₀2
= 0.47712 + 1 - 0.30103 = 1.17609


Example 5:

If log₅ = 0.6990, and 10ʸ = 0.05, Find the value of y

Solution:
10ʸ = 0.05 ⇒ take log of both side
log10ʸ = log0.05

y log10 = log0.05

y log10 = log5/100
ylog10 =log5 - log10²

y x1 = 0.6990 - 2log10

y = 0.6990 - 2 x1

y = 0.6990 - 2



Example 6:

Simplify each of the following
a.  log₆₄0.125    

 b.   log₁₀25 +      

       log₁₀32 - log₁₀8
     



Solutions:


(a) log₆₄0.125

  =  log0.125/log64

=  log125/1000 ÷ log64 

log8 ÷ log64
log8/log64

log2³/log2⁶

3log2/6log2

log2 cancled log2
=3/6 = 1/2


(b).  log₁₀25 +log₁₀32 +log -log8

log₁₀(25 x 32 )/8

log₁₀100 = log₁₀10² = 2log₁₀10 = 2 x1 = 2

Example 7:

If log₆4=z, log₅6 = y, Express the following in term of x and y

i.   log₆64


solution

i.    log₆64 = log₆4³ = 3log₆4 = 3 x z= 3z




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