数学 - Python - 急速・緩慢に変化する関係 - 指数関数・対数関数 - 指数関数と対数関数 - 対数 - 真数、底数、等式、解

学習環境

新装版数学読本2 (松坂和夫(著)、岩波書店)の第7章(急速・緩慢に変化する関係 - 指数関数・対数関数)、7.2(指数関数と対数関数)、対数の問14の解答を求めてみる。

1. $\begin{array}{l} 3^{2} = x \\ x = 9 \end{array}$
2. $x = 4^{\cdot 1} = \frac{1}{4}$
3. $\begin{array}{l} x = 8 1^{\frac{1}{4}} \\ = {(3^{4})}^{\frac{1}{4}} \\ = 3 \end{array}$
4. $\begin{array}{l} x = {(0.2)}^{- 2} \\ = {(\frac{1}{5})}^{- 2} \\ = 25 \end{array}$
5. $\begin{array}{l} x^{2} = 25 \\ x^{2} = 5^{2} \\ x = 5 \end{array}$
6. $\begin{array}{l} x^{- 3} = \frac{1}{1000} \\ = 1 0^{- 3} \\ x = 10 \end{array}$
7. $\begin{array}{l} x^{4} = 2^{8} \\ x^{4} = {(2^{2})}^{4} \\ x = 4 \end{array}$
8. $\begin{array}{l} x^{- 2} = 4 \\ x^{- 2} = 2^{2} \\ x^{- 2} = \frac{1}{2^{- 2}} \\ x^{- 2} = {(\frac{1}{2})}^{- 2} \\ x = \frac{1}{2} \end{array}$
9. $\begin{array}{l} x^{\frac{1}{4}} = \sqrt{3} \\ = 3^{\frac{1}{2}} \\ = {(3^{2})}^{\frac{1}{4}} \\ x = 9 \end{array}$

#!/usr/bin/env python3 from sympy import pprint, symbols, sqrt, root, log, Rational, solve from unittest import TestCase, main print('14.') class MyTestCase(TestCase): def setUp(self): pass def tearDown(self): pass def test(self): x = symbols('x') eqs = [log(x, 3) - 2, log(x, 4) + 1, log(x, 81) - Rational(1, 4), log(x, Rational(1, 5)) + 2, log(25, x) - 2, log(Rational(1, 1000), x) + 3, log(256, x) - 4, log(4, x) + 2, log(sqrt(3), x) - Rational(1, 4)] egg = [9, Rational(1, 4), 3, 25, 5, 10, 4, Rational(1, 2), 9] for s, t in zip(eqs, egg): self.assertEqual(solve(s), [t]) if __name__ == '__main__': main()

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Kamimura's blog

2019年9月9日月曜日

数学 - Python - 急速・緩慢に変化する関係 - 指数関数・対数関数 - 指数関数と対数関数 - 対数 - 真数、底数、等式、解

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