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Logarithms. This is pronounced as log to the base 2 of x. log 4. 64. =. 3. log 2. 32. =. 5. log 8. p. =. 2. b). Express the following in exponential form:. i). log 2 8 = 3. i). log 2 1 / 128 = -7. 1 / 128. 8. =. 2 3. =. 2 -7. Examples. a).
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log4 64 = 3 log2 32 = 5 log8 p = 2 b) Express the following in exponential form: i) log28 = 3 i) log21/128 = -7 1/128 8 = 23 = 2-7 Examples a) Write the following in logarithmic form: i) 43 = 64 ii) 25 = 32 iii) p = 82 (Check: 2 × 2 × 2 = 8)
Evaluate: a) log 2 8 b) log 3 1 c) log 2 1/16 Laws of Logarithms Law A: log a a n = n
Law B: Law C: Law D:
Logarithmic Equations Rules learned can now be used to solve logarithmic equations. log 3 x + log 3x3 – (log 3 8 + log 3x) = 0 log 3 x + log 3x3 – log 3 8 - log 3x = 0 log 3x3 – log 3 8 = 0 log 3x3 = log 3 8 x3 = 8 x = 2
2) log 4 x + 2 log 4 3 = 0 3) 32x-7 = 243 35 = 243 log 4 x + log 4 32 = 0 log 4x = - log 4 32 5 = 2x - 7 log 4x = log 4 3-2 2x - 7 = 5 x = 3 -2 2x = 12 x = 1/32 x = 6 x = 1/9
Warm up 11/2 Simplify:
Natural Logarithms Logarithms to the base e are called natural logarithms. log ex = ln x i.e. log 2.718x = ln x Examples Solve the following, rounding correct to 4 d.p.: a) lnx = 5 b) e x = 7 (Take natural logs of each side) log ex = 5 ln e x = ln 7 xlogee = ln 7 x = e 5 x = 148.41316 x = ln 7 x = 148.4131 (4dp) x = 1.9459101 x = 1.9459 (4dp)
c) 37x+2 = 30 d) For the formula P(t) = 50 e -2t ln 3 7x+2 = ln 30 i) Evaluate P(0). (7x+2) ln 3 = ln 30 ii) For what value of t is P(t) = ½ P(0)? 7x+2 = ln 30/ln 3 i) P(0) = 50 e –2(0) 7x+2 = 3.0959033 = 50 e 0 7x = 1.0959033 = 50 x = 0.1565576 ii) ½ P(0) = 25 50 e -2t = 25 x = 0.2 (1dp) e -2t = ½ Remember: ln e = loge e ln e -2t = ln ½ -2t = -0.6931471 t = 0.3465735 t = 0.3 (1dp)
y x Formula from experimental data • If data from an experiment is analysed, say x and y, and plotted, and it is found to form a straight line then x and y are related by the formula: y = mx + c • If data from an experiment is analysed, say x and y, and plotted, and it is found to form an exponential growth curve then x and y are related by the formula: y = k xn (k and n are constants)
Using logarithms y = k xn can be expressed as a straight line. y = k xn log 10 y = log 10 k xn log 10 y = log 10 k + log 10xn log 10 y = log 10 k + n log 10x log 10 y = n log 10x + log 10 k Y = mX + C Where: Y = log 10 y If y = k xn X = log 10x C = log 10 k Then log 10 y = n log 10x + log 10 k n = gradient