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
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 : loga(m/n) = loga m - loga 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 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
22
= 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
77 (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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