Understanding Logarithms and Logarithmic Laws
Logarithms help us determine the power to which a base must be raised to give a specific result. This content explains concepts like logarithmic laws, changing bases, solving logarithmic equations, and includes examples for better understanding.
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3C Logarithms the base 10 raised to the power 2 gives 100 102 = 100 2 is the power which the base 10 must be raised to, to give 100 the power = logarithm 2 is the logarithm to the base 10 of 100 Logarithm is the number which we need to raise a base to for a given answer to what power must I raise 2 to give an answer of 64? ans = 6 written as log2 64 = 6
to what power must I raise 2 to give an answer of 64? ans = 6 written as log2 64 = 6 to what power must I raise 5 to give an answer of 625? ans = 4 written as log5 625 = 4 to what power must I raise 9 to give an answer of 3? ans = 1 2 written as log9 3 = 1 2 loga n = p ap = n baseanswer = number inside
1. loga m + loga n = loga mn 2. loga m - loga n = loga m 3. n loga m = loga mn n 4. logn m = loga m loga n change of base law N.B. The log of a negative is impossible to find
Law 2 loga m - loga n = loga m Let loga m = p & loga n = q ap = m ap = aq ap - q =m n m n m n Proofs: Law 1 loga m + loga n = loga m n Let loga m = p & loga n = q ap = m ap . aq = m . n ap + q = m . n loga m . n = p + q loga m n = loga m + loga n n aq = n aq = n m n loga = p - q baseanswer = number inside loga = loga m - loga n
Law 3 n loga m = loga mn Law 4 logn m = loga m loga n Let loga m = p Let logn m = p ap = m We need mn take logs of both sides m np = ( )n = m ( )n ap apn = mn loga mn = pn loga mn = (loga m) n loga mn = n loga m loga np = loga m p loga n = loga m p = loga m baseanswer = number inside loga n logn m = loga m loga n
e.g.1 log4 64 = x e.g.3 log4 (5x + 6) = 2 baseanswer = number inside 42= 5x + 6 16= 5x + 6 10= 5x 10 5 2 = x 4x= 64 4x= 43 x = 3 = x e.g.2 log2 x = 5 25= x x = 32
e.g.4 log3 (2x - 4) = 1 + log3 (4x + 8) log3 (2x 4) log3 (4x + 8) 2 4 log 4 8 x + 2 4 3 4 8 x + 3(4 8) 2 4 x x + = For an unknown power always take logs of both sides e.g.5 6n = 3200 = 1 = x 1 log10 6n = log10 3200 3 n log10 6 = log10 3200 x = 1 n = log10 3200 log10 6 n = 4.5045 12x + 24 = 2x - 4 10x = -28 x = -28 x = -2.8 10
Calculations using log10 logn m = logam change of base law loga n log10 1000 = 3 as 103 = 1000 If we want log2 32 = log10 32 log10 2 e.g.6 log2 55 = log10 x log10 55 log10 2 = log10 x baseanswer = number inside 5.78 = log10 x 105.78 = x x = 604449
