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Logarithms for Analog Circuits
16 questions By Tony R. Kuphaldt
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Question 16 of 16
Logarithms have interesting properties, which we may exploit in electronic circuits to perform certain complex operations. In this question, I recommend you use a hand calculator to explore these properties.
Calculate the following:
- 10log3 =
- log(108) =
- eln3 =
- ln(e8) =
- 10(log3 + log5) =
- e(ln3 + ln5) =
- 10(log2.2 + log4) =
- e(ln2.2 + ln4) =
- 10(log12 − log4) =
- e(ln12 − ln4) =
- 10(2 log3) =
- e(2 ln3) =
- 10([log25/2]) =
- e([ln25/2]) =
Reveal answer- 10log3 = 3
- log(108) = 8
- eln3 = 3
- ln(e8) = 8
- 10(log3 + log5) = 15
- e(ln3 + ln5) = 15
- 10(log2.2 + log4) = 8.8
- e(ln2.2 + ln4) = 8.8
- 10(log12 − log4) = 3
- e(ln12 − ln4) = 3
- 10(2 log3) = 9
- e(2 ln3) = 9
- 10([log25/2]) = 5
- e([ln25/2]) = 5
Notes:Discuss what mathematical operations are being done with the constants in these equations, by using logarithms. What patterns do your students notice? Also, discuss the terms “log” and “antilog,” and relate them to opamp circuits they’ve seen.
Ask your students whether or not they think it matters what “base” of logarithm is used in these equations. Can they think of any other arithmetic operations to try using logarithms in this manner?
