F.4 Maths ( 2 )

對不起我習慣用中文作答。
Energy E ( J ) transmitted from a source through a medium is given by the
following equation:
E= a‧6bt
where a,b are constants and t ( s ) is the time of transmission.
a)Initially,the energy transmitted from the source is 500J .Find the value of a.
(ans:500)
你的公式應為
E = a 6bt
開始時 t = 0，E = 500J
所以
E = a 6bt
500 = a 6b x 0
a = 500J


b)After 2 s,the energy transmitted is dropped by 200 J,find the value of b.
(ans:-0.143)
t = 2，E = 500 – 200 = 300J
則照代公式
E = a 6bt
300 = 500 x 62b
0.6 = 62b
log(6) 0.6 = 2b (log(6) 0.4 = log0.4 / log 6)
b = log 0.6/(2log6)
b = -0.143s-1

c)Find the time taken when the energy transmitted is dropped to half of the
initial value. (ans: 2.71s)
減少一半即 E = 250J
將資料照代公式
E = a 6bt
250 = 500 x 6-0.143t
0.5 = 6-0.143t
log(6)0.5 = -0.143t
log0.5 / log 6 = -0.143t
-0.387 = -0.143t
t = 2.71s

d)An instrument is used to detect the energy level.However,the enery
level below 1x10^-9 J could not be detected.How long will it take the
reading of the instrument to drop to zero? (ans: 106s )
當 E = 1x10-9J 求 t
照代公式
E = a 6bt
1x10-9 = 500 x 6-0.143t
2x10-12 = 6-0.143t
log(6)2x10-12 = -0.143t
log2x10-12 / log 6 = -0.143t
t = 105s (因為尾數問題所以和答案相差少許)

註 log(x)y 表示以 x 為底的對數
log(x)y = log y / log x
因為指數的逆運算等於對數
所以
xz = y
z = log(x)y = log x / log y