Evaluating Logarithm

In this section, student will learn evaluating logarithm.
The word logarithm was coined from two Greek words"logos" which means a ratio and "arithmos" means numbers.
Logarithms and exponents are closely related to each other. The inverse of these operations gives us two distinct operations viz. extracting roots and taking logarithms.
am = b is exponential form.

It is read as mth power of 'a' is 'b'.
For each positive real number a, a≠ 1, the unique real number 'm' is called the logarithm of 'b' to the base 'a' or
logab = m if and only if am = b

'log' being the abbreviation of the word 'logarithm.

Properties of logarithm
1) loga1 = 0 because a0=1
2) logaa = 1 because a1=a
3) logaax = x because ax =ax

Evaluating logarithm

1) log327
Solution : We know from definition that
logab = m, if and only if am = b, a > 0 and a≠1
Let m = log327
∴ 3m = 27
3m = 33
∴ m = 3
so, log327= 3

2) log2√(32)
Solution : We know from definition that
logab = m, if and only if am = b, a > 0 and a≠1
Let m = log2√(32)
∴ 2m = √(32)
2m = (25)1/2
2m = 25/2
∴ m = 5/2
so, log2√(32)= 5/2

Exponential to logarithmic form
27 = 128 ⇒ log 2128 = 7
34 = 81 ⇒ log 381 = 4
60 = 1 ⇒ log 61 = 0
a-b = c ⇒ log ac = -b
43 = 64 ⇒ log 4 64 = 3
72 = 49 ⇒ log 7 49 = 2
53 = 125 ⇒ log 5 125 = 3
103 = 1000 ⇒ log 10 1000 = 3
83 = 512 ⇒ log 8 512 = 3
15-2 = 1/225 ⇒ log 15 (1/225) = -2
Logarithmic to Exponential form
log2 1 = 0 ⇒ 20 = 1
log525 = 2 ⇒ 52 = 25
log7343 = 3 ⇒ 73 = 343
log10100 = 2 ⇒ 102 = 100
log816 = 4/3 ⇒ 84/3 = 16


11th grade math

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