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) log
a1 = 0 because a
0=1
2) log
aa = 1 because a
1=a
3) log
aa
x = x because a
x =a
x
Evaluating logarithm
1) log
327
Solution : We know from definition that
log
ab = m, if and only if a
m = b, a > 0 and a≠1
Let m = log
327
∴ 3
m = 27
3
m = 3
3
∴ m = 3
so, log
327= 3
2) log
2√(32)
Solution : We know from definition that
log
ab = m, if and only if a
m = b, a > 0 and a≠1
Let m = log
2√(32)
∴ 2
m = √(32)
2
m = (2
5)
1/2
2
m = 2
5/2
∴ m = 5/2
so, log
2√(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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