FINDING LOGARITHMS WITHOUT USING TABLE
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

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