指數與對數(7)
Author: 藍邦偉　老師

∵1=2×5^{2k}⇒5^{2k}=\frac{1}{2}

$ ⇒2k=log_5\frac{1}{2} $

$ ⇒k=\frac{1}{2}log_5\frac{1}{2}=log_5\left( \frac{1}{2}  \right)^{\frac{1}{2}}=log_52^{-\frac{1}{2}}  $

$ ∴0.01>2×5^{kx} $

$ ⇒log0.01>log(2×5^{kx}) $

$ ⇒-2>log2+(kx)log5$

$ ⇒ -2-0.3010>x(log5)(log_52^{-\frac{1}{2}})=xlog2^{-\frac{1}{2}}$

$ ⇒-2.3010>(-\frac{1}{2}log2)x $

$⇒ 2.3010<\frac{0.3010}{2}x $

$ ⇒x>\frac{4.6020}{0.3010}=15.3 $

$ ⇒x=16 $

16

1

2

假設喝完酒x小時後酒精濃度正好降到0.01％

$ \left\lbrace \begin{array}{ll} 1=2×5^{2k}  \ 0.01=2×5^{kx} \ \end{array}  \right.\ $

$ ⇒\left\lbrace\begin{array}{ll} 1=\left(  2×5^{2k} \right)^x  \ 0.01²=\left( 2×5^{kx}  \right)²  \ \end{array}   \right.\  $

⇒\frac{0.0001}{1}=\frac{2^2}{2^x}=2^{2-x}

⇒log0.0001=log2^{2-x}=(2-x)log2

⇒-4=(2-x)×0.3010

⇒x-2=\frac{4}{0.3010}=13.2

⇒x=15.2

$ 取x=16 $

n = 1 (slider from 1 to 13, step 1)
A = (-1.06, 2.56)
B: A + (0.7, 0)
D: A + (0, 0.4)
C: B + D - A
poly1, a, b, c, d = Polygon(A, B, C, D)
e = true
f = false
k = 10 (slider from 1 to 10, step 1)