17 Feb 2008 | 2 min. (257 words)

study material-mathematics-algebra-logarithms

Logarithms - 2

log_a(m/n) = log_am - log_an
Why ?
Let log_am = x
      log_an = y
m = a^x and n = a^y [Using definition]
     m/n = a^x/a^y = a^{x - y}
=> log_am/n = x - y
                  = log_am - log_an
log_amⁿ = n log_am.
why ?
Let log_amⁿ = y
By definition a^y = mⁿ
              or, a^y/x = m
Again using definition of log
       log_am = y/x
or n log_am = y
so, log_amⁿ = n log_am.

Illustration - 2. If log₁₂27 = a, then log₃16 = ??
log₃16 = log₃2⁴ = 4 log₃2 ……………………………………….. (i)
Now, log₁₂27 = log₁₂3³
                      = 3 log₁₂3
                      =
So, log₃2 =
Now, log₃ =

*7. log = log_an
log = p log_aqⁿ - (Using formula 6.)
             = p - (Using formula 3.)
             = - (Using formula 6.)
             = - (Using formula 3.)

*8. = n

Illustration - 3.
Evaluate.

          = (Using formula 7)

(Using formula 6)
= 9 (using formula 8)
Hence = 9

      log_pa > log_pb
=> a b if p is greater than 1 i.e. p > 1
or b a if p is positive and less than 1, 0 < p < 1

Remembering Tip:- If base if > 1 then inequality remains same and if base is + ve but less than 1 then sign of inequality is reversed.
Why ?
Let log_pa = x
      log_pb = y
So, a = p^x
and b = p^y
Now x > y is given
So, if p > 1
then a b
and if 0 < p < 1
then a b

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