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 loga m = x and logan = y
ax = m and ay = n
∴ mn = ax . ay
mn = ax + y
By definition of logarithm
loga(mn) = x + y
loga(mn) = loga m + loga 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 loga m = x and logan = y
ax = m and ay = n
∴ m/n = ax / ay
m/n = ax - y
By definition of logarithm
loga(m/n) = x - y
loga(m/n) = loga m -loga 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 : loga(m)n = n loga m
Proof : Suppose loga m = x
ax = m
(ax)n = (m)n ( taking nth power on both sides)
By definition of logarithm
loga(m)n = n. x
loga(m)n = n loga 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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