COMMON LOGARITHMS -- DEFINITION, SOLVED EXAMPLES, EXERCISES, LINKS TO FURTHER STUDY

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Definition of Common logarithm :

The base 10 of a logarithm has a significance.

In Experimental Science and Engineering,
the results of the investigation are presented in graphs.
How one variable varies with the other is studied by
plotting the variables on X Axis and Y Axis.

Some times it is useful to take one parameter
on one Axis (usually X-Axis) with logarithmic scale,
the other being ordinary scale.

Such a plot is called Semi-Log plot.

Some times, we also take the parameters on both Axes
to logarithmic scale which is called Log-Log plot.

In all these cases the base of the logarithm is
invariably 10.

The Logarithms with base 10 are called
Common Logarithms.

Here, In Common Logarithms, we give some
Solved Examples and Exercise in Logarithms.

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Solved Example 1 of Common Logarithms :

Example 1 of Common Logarithms :

(i) If log12 27 = a, compute log6 32 in terms of a
(ii) If log30 3 = a and If log30 5 = b, find log30 8

Solution to Example 1 of Common Logarithms :

Solution to 1(i) of Common Logarithms :

If log12 27 = a, compute log6 32
We have a = log12 27 = log12 33
We know, in log of a power (See Formula 7), the exponent multiplies the log.
a = 3 log12 3
We know, log of a quantity to a base can be written asthe ratio of log of the quantity and log of the base.(See Formula 8)
a = 3 {(log 3)⁄(log 12)}
We know 12 = 3 x 4 = 3 x 22 ⇒ log 12 = log (3 x 22)
We know, log of a product is equal to the sum of the logs of the factors of the product (See Formula 5)
∴ log 12 = log 3 + log (22)
We know, in log of a power (See Formula 7), the exponent multiplies the log.
∴ log (22) = 2 log 2
Thus log 12 = log 3 + 2 log 2
a = 3 {(log 3)⁄(log 3 + 2 log 2 )}
Dividing numerator and denominator by log 3, we get
a = 3 [(1)⁄{1 + 2 (log 2⁄log 3)}]
= 3⁄(1 + 2p) where p = (log 2⁄log 3)
⇒ 1 + 2p = 3⁄a ⇒ 2p = 3⁄a - 1 = (3 - a)⁄a
p = (3 - a)⁄2a ...........(i)

Let x = log6 32
We know, log of a quantity to a base can be written asthe ratio of log of the quantity and log of the base.(See Formula 8)
x = (log 32)⁄(log 6)
We know 32 = 2 x 2 x 2 x 2 x 2 = 25; 6 = 2 x 3;
x = {log (25}⁄{log (2 x 3)}
Numerator is logarithm of a power and Denominator is logarithm of a product.
We know, in log of a power (See Formula 7), the exponent multiplies the log.
We know, log of a product is equal to the sum of the logs of the factors of the product (See Formula 5)
x = {5 log 2}⁄{log 2 + log 3}
Dividing numerator and denominator by log 3, we get
x = 5 (log 2⁄log 3)⁄{(log 2⁄log 3) + 1}
= 5p⁄(p + 1) [Since p = (log 2⁄log 3)]
We have p = (3 - a)⁄2a ...........(i)
p + 1 = (3 - a)⁄2a + 1 = (3 - a + 2a)⁄2a = (3 + a)⁄2a........(ii)
Substituting the value of p from (i) and (p + 1) from (ii) in x, we get
x = 5 {(3 - a)⁄2a}⁄{(3 + a)⁄2a} = 5 {(3 - a)⁄(3 + a)}. Ans.

Solution to 1(ii) of Common Logarithms :

we know 30 = 3 x 5 x 2; log30 30 = 1
∴ log30 30 = 1 ⇒ log30 (3 x 5 x 2) = 1
We know, log of a product is equal to the sum of the logs of the factors of the product (See Formula 5)
∴ log30 3 + log30 5 + log30 2 = 1
a + b + log30 2 = 1 ⇒ log30 2 = 1 - a - b
log30 8 = log30 (23)
We know, in log of a power (See Formula 7), the exponent multiplies the log.
∴ log30 8 = 3(log30 2) = 3(1 - a - b). Ans.





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Solved Example 2 of Common Logarithms :

Example 2 of Common Logarithms :

If b = √(ac), prove that
(loga n)⁄(logc n) = {(loga n) - (logb n)}⁄{(logb n) - (logc n)}

Solution to Example 2 of Common Logarithms :

By data b = √(ac) ⇒ b2 = ac ⇒ ba = cb.........(i)
Let x = logn a then 1⁄x = 1⁄logn a;
Let y = logn b then 1⁄y = 1⁄logn b;
Let z = logn c then 1⁄z = 1⁄logn c;
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ 1⁄x = loga n; 1⁄y = logb n; 1⁄z = logc n
To prove (loga n)⁄(logc n) = {(loga n) - (logb n)}⁄{(logb n) - (logc n)}
L.H.S. = (loga n)⁄(logc n) = (1⁄x)⁄(1⁄z) = zx ......(ii)
R.H.S. = {(loga n) - (logb n)}⁄{(logb n) - (logc n)}
= {(1⁄x) - (1⁄y)}⁄{(1⁄y) - (1⁄z)} = {(y - x)⁄(xy)}⁄{(z - y)⁄(yz)}
= {(y - x)⁄(xy)} x {(yz)⁄(z - y)} = (zx){(y - x)⁄(z - y)}.........(iii)
(y - x)⁄(z - y) = (logn b - logn a)⁄(logn c - logn b)
We know, log of a quotient can be written as the difference of the logs of the numerator and denominator of the quotient (See Formula 6)
Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.
∴ (y - x)⁄(z - y) = {logn (ba)}⁄{logn (cb)} = 1 [Since ba = cb from (i)]
Using this value in (iii), we get
R.H.S. = (zx) (1) = zx
From (ii), L.H.S. = zx
∴ L.H.S. = R.H.S. (Proved.)

Thus Example 2 of Common Logarithms is solved.



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More Solved Examples

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Exercise : Common Logarithms

Exercise Problems : Common Logarithms

  1. If 1⁄x = 1 + loga bc, 1⁄y = 1 + logb ca, 1⁄z = 1 + logc ab,
    then prove that x + y + z = 1.
  2. If a = log24 12, b = log36 24, c = log48 36,
    then prove that 1 + abc = 2bc.


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