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 : log_a(mn) = log_a m + log_a n
Proof : Suppose log_a m = x and log_an = 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 : log_a(m/n) = log_a m - log_a n
Proof : Suppose log_a m = x and log_an = 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 log₂ 16 - log₂ 8
Solution : Consider log₂ 16 - log₂ 8
log₂ 16 - log₂ 8 = log₂(16/8) --------> By 2nd rule of logarithms
= log₂2
= 1 ( By properties of logarithm)

3rd rule : log_a(m)ⁿ = n log_a 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 log₆ 36
Solution : log₆ 36
= log₆ 6²
= 2 log₆ 6 ( By 3rd rule)
∴ log₆ 36 = 2

Solve for x :
1) x = log ₇ 343
x = log ₇ 7³
x = 3 log₇7 (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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