Rules for Logarithms

The following are the rules for logarithms are nothing but power laws of exponents which are studied in earlier classes. These rules hold for any base 'a' (a > 0 and a ≠ 1 )

Rules for logarithms

1st rule : loga(mn) = loga m + loga n
Proof : Suppose log a m = x and log a n = y
a
x = m and a y = n
∴ mn = a
x . a y
mn = a
x + y
By definition of logarithm
log
a (mn) = x + y
log
a (mn) = log a m + log a n

Example : Prove that : log 12 = log 3 + log 4
Solution : Consider log 3 + log 4
log (3 x 4) --------> By 1st rule of logarithms
= log 12
∴ log 12 = log 3 + log 4

2nd rule : loga(m/n) = loga m - loga n
Proof : Suppose log a m = x and log a n = y
a
x = m and a y = n
∴ m/n = a
x / a y
m/n = a
x - y
By definition of logarithm
log
a (m/n) = x - y
log
a (m/n) = log a m -log a n

Example : Find log 2 16 - log 2 8
Solution : Consider log 2 16 - log 2 8
log
2 16 - log 2 8 = log 2 (16/8) --------> By 2nd rule of logarithms
= log
2 2
= 1 ( By properties of logarithm)

3rd rule : loga(m)n = n loga m
Proof : Suppose log a m = x
a
x = m
(a
x ) n = (m) n ( taking nth power on both sides)
By definition of logarithm
log
a (m) n = n. x
log
a (m) n = n log a m

Example : Find log 6 36
Solution : log 6 36
= log
6 6 2
= 2 log
6 6 ( By 3rd rule)
∴ log
6 36 = 2

Solve for x :
1) x = log
7 343
x = log
7 7 3
x = 3 log
7 7 (by 3rd rule)
x = 3 x 1
x = 3

2) log 2 + log(x+2) - log (3x-5) = log 3
Solution : log 2 + log(x+2) - log (3x-5) = log 3
By 1st rule
log 2(x+2) - log (3x-5) = log 3
By 2nd rule log 2(x+2)/(3x-5) = log 3
2(x+2)/(3x-5) = 3/1
2(x+2) = 3(3x-5) -----(cross multiply)
2x + 4 = 9x -15
2x -9x = -15-4
-7x = -19
∴ x = 19/7

11th grade math

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