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Logarithms 102 = 100 the base 10 raised to the power 2 gives 100

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Presentation on theme: "Logarithms 102 = 100 the base 10 raised to the power 2 gives 100"— Presentation transcript:

1 Logarithms 102 = 100 the base 10 raised to the power 2 gives 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

2 loga n = p ap = n baseanswer = number inside
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

3 loga m + loga n = loga mn 2. loga m - loga n = loga m n
n loga m = loga mn logn m = loga m change of base law loga n N.B. The log of a negative is impossible to find

4 baseanswer = number inside
Proofs: Law 1 Law 2 loga m - loga n = loga m n loga m + loga n = loga m n Let loga m = p & loga n = q Let loga m = p & loga n = q ap = m aq = n ap = m aq = n ap . aq = m . n ap = aq m n ap + q = m . n loga m . n = p + q m n ap - q = loga m n = loga m + loga n m n loga = p - q baseanswer = number inside m n loga = loga m - loga n

5 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 np = m ( )n ap = ( )n m loga np = loga m apn = mn baseanswer = number inside p loga n = loga m loga mn = pn p = loga m loga n loga mn = (loga m) n loga mn = n loga m logn m = loga m loga n

6 baseanswer = number inside
e.g log4 64 = x e.g log4 (5x + 6) = 2 baseanswer = number inside x 42 = 5x + 6 4 = 64 16 = 5x + 6 4x = 43 10 = 5x x = 3 10 5 = x e.g log2 x = 5 2 = x 25 = x x = 32

7 e.g log3 (2x - 4) = 1 + log3 (4x + 8) For an unknown power always take logs of both sides log3 (2x – 4) – log3 (4x + 8) = 1 e.g n = 3200 log10 6n = log n log10 6 = log n = log log10 6 n = 12x + 24 = 2x - 4 10x = -28 x = -28 10 x = -2.8

8 Calculations using log10
logn m = loga m change of base law loga n log = 3 as 103 = 1000 If we want log2 32 = log10 32 log10 2 e.g log2 55 = log10 x log10 55 = log10 x log10 2 baseanswer = number inside 5.78 = log10 x = x x =


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