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IB Mathematics: Applications and Interpretation HL · 12.6: Exponents & Logarithms · 12.6.2
Logarithms Problem set A Answers
Skills
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- (a) \(x = \frac{1}{1000}\)
- (b) \(y = \frac{401}{100}\)
- (c) \(t = \frac{9}{2}\)
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- (a) \(x = e^3 = 20.1\)
- (b) \(x = e^{-2.5} = 0.0821\)
- (c) \(x = \frac{e^2+3}{4} = 2.60\)
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- (a) \(5 p - 2 q\)
- (b) \(\frac{2 p}{3} + \frac{q}{3}\)
- (c) \(\frac{p}{2} - 3 q + 3\)
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- (a) \(x = \frac{25}{2}\)
- (b) \(x = 16\)
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- (a) \(x = 4\)
- (b) \(x = \frac{e^3+1}{e^3-2} = 1.17\)
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- (a) \(x = \frac{2\ln 3 + \ln 4}{\ln 4 - \ln 3} = 12.5\)
- (b) \(x = \frac{\ln 2}{\ln 7 - 3\ln 2} = -5.19\)
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- (a) \(x = −1.40\)
- (b) \(x = 33\)
- (c) \(x = 6.31\)
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- (a) \(x = 3.40\)
- (b) \(x = e^3 = 20.1\)
- (c) \(x = 33.1\)
In context
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- (a) \(\text{pH} = 4.52\)
- (b) \(h = 6.31 \times 10^{-5}\)
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- (a) \(H = 5.45\) m
- (b) \(t = e^2 = 7.39\) years
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\(t = \frac{\ln 1.5}{\ln(1.03/0.99)} = 10.2\) years
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- (a) \(M = 5.30\)
- (b) \(A = 31 600\) micrometres
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- (a) \(C = 53.0\%\)
- (b) \(t = e^2 = 7.39\) hours
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- (a) b = 1.32, a = 2.55
- (b) D = 7.71 cm
- (c) t = 62.4 years
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- (a) \(E=6.31\times10^{13}\) J
- (b) M = 7.00
- (c) 1000 times
Stretch
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\(x = 5\)
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\(x = \frac{\ln 15}{\ln 1.8} = 4.61\)
jmaths.xyz · AI HL · 12.6.2 Logarithms and solving exponential and log equations