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Logarithm

We know that a² ₌ b is n exponential and has the meaning b = a ´a.

Similarly a^1/2= b means b = square root of a. They can b expressed in logarithmic form as follows:

a² ₌ b Þ log _a b = 2; read as logarithm of b is 2 ; that is logarithm of b to the base a is the index 2.

Hence logarithm expresses the index of the base a.

In the case of a^1/2= b it is log _a b= ½

Thus any exponential expression can be converted to logarithmic form.

All the laws, lemma, of exponential help us to apply logarithm and use it conveniently for calculation purpose.

Here is a table of properties of both exponential and logarithmic expressions:

S.No	Exponential	Logarithmic
1	a^2 = b	log a b = 2
2	a ^m ´ a ^n = a ^(m+n)	Log a (a ^m ´ a ^n)= m + n
3	a ^m ¸ a ^n = a ^(m --n)	Log a (a ^m ¸ a ^n)=m – n
4	(a ^m) ^n = a ^mn	Log a (a ^m) ^n = n log (a ^m) =nm

Let us see how these laws are being used:

Consider the log product 32 x 64 ; using base 2 we find log(32 x 64) = ?

Answer:

Log₂ (32 x 64) = log (2 ⁵ ´ 2 ⁶) = 5 + 6 = 11. the base is 2.

Now let us find:

The log of ( 32 ¸64) = 5 – 6 = -1 ( again base is 2)

How is logarithm useful for an applied mathematician, engineers, and scientists?

These professionals use logarithm widely to solve complicated multiplications, divisions, n^th roots and mix of all these. They either use base 10 (called common logarithm) or base e (Napier’s base). Logarithm with base e is called natural logarithm.

There is a ready reckoner table of logarithms containing log ₁₀ (from 0.0000 to 99.9999).

How far is the accuracy? Values calculated using 4 decimal log table is as accurate as the 4 decimals are. But the above professional (mostly scientist, Technocrats use a special log table which is 7 decimal place accurate!)

Here are some properties of Logarithms:

Laws of logarithms:

For m,n >0, a>0 and a¹1

1. log _a1=0 , log _aa=1.
2. log_a (mn)=log_a m +log_a n ; log _a(m₁ m₂ … m_n)= log m₁ + logm₂+…+logm_n
3. log _a()=log _a m –log _a n
4. log _a (mⁿ) = n log _am
5. log _aⁿ m = 1/n log _a m
6. log _a m = and log _ba log _ab =1
7. a^{log b} = b ^{log a}; a,b >0.
8. If a>1, then x > y > 0 Û log _ax > log _ay and if 0<a<1 then x>y>0 Û log x< logy

Examples:

1. Evaluate log ₆₂₅ 15625.

15625 =5 ⁶ and 625 = 5 ⁴ then log ₆₂₅ 15625 = = = 6/4=3/2

2. log= ½(log x+log y), show that x=y.

Solution: log= ½(log x+log y) =½ log xy = log (xy) ^½

= (xy) ^½ ; squaring both sides we get ()² = xy Þ x²+y²+2xy=4xy

Þ x²+y²-2xy =0 Þ(x – y)²=0 i.e x=y.

3. Prove that x ^logy-logz ´ y ^{logz – log x} ´ z ^logx-logy = 1 .

Solution: taking logarithm on both sides we get:

If Log(x ^logy-logz ´ y ^{logz – log x} ´ z ^logx-logy) = log 1 then (logy logz)logx +(logz-logx)logy +(logx – logy)logz=0 and log 1=0 hence the result.

4. Let x, y, z be three distinct positive numbers and then show that (i) xyz=1 (ii) x^x y^y z^z =1

Solution: Given =k (say)

Logx =k(y-z), logy =k(z-x) ; logz =k(x-y) Adding we get

Logx+logy+logz = k(y-z+z-x+x-y)=0

Log xyz=0 implies xyz=1

Also xlogx =kx(y-z) ; ylogy=kyz-x) ; zlogz=kz(x-y)

Adding these we get: xlogx+ylogy+zlogz=k(0) =0 hence log x^x y^y z^z=0 thus x^x y^y z^z=1

5 Evaluate: 0.5

Solution: 0.5 = 0.5=0.5 = (1/2)

= = 1/1/9=9.

5. Using log 2=0.30103 and log 3=0.47712 show that (1)¹⁰⁰⁰ > 100000.

Solution: Taking log. Log (1)¹⁰⁰⁰ = 1000 log (81/80) = log ()

1000(4log3 – 3 log2 – log10) = 1000( 4 x 0.47712 – 3 x 0.30103 –1)

= 1000 x 0.00539 = 5.39 > 5. Hence the answer.

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