標題:
Form4 maths question about logarithmic functions(3)
發問:
40.Consider 2x=5y=10z, where x,y and z are non-zero real numbers.(a) Express x,y in terms of z respectively.(b) Hence show that 1/x+1/y=1/z41.If x0.25 +x(-0.25)=4, find(a)(√x )+(1/√x)(b)x+(1/x)44.The population of a city is four million at the end of the year 204. The population grows... 顯示更多 40.Consider 2x=5y=10z, where x,y and z are non-zero real numbers. (a) Express x,y in terms of z respectively. (b) Hence show that 1/x+1/y=1/z 41.If x0.25 +x(-0.25)=4, find (a)(x )+(1/x) (b)x+(1/x) 44.The population of a city is four million at the end of the year 204. The population grows at the rate 3% per year. (a) In which year will the population be doubled? Ans: 40(a)x= z/log2 y=z/log5 41.(a)14 (b)194 44.(a)2028 唔該列式教教我點計,多謝!
最佳解答:
Q1. 2x = 10z x log 2 = z log 10 = z........(1) 5y = 10z y log 5 = z log 10 = z .........(2) So 1/x + 1/y = log 2/z + log 5/z = (log 2 + log 5)/z = (log 10)/z = 1/z. Q2. [x0.25 + x(-0.25)]2 = x0.5 + 2(x0.25)[x(-0.25)] + x(-0.5) = x0.5 + 2 + x(-0.25) = 42 = 16, therefore, x0.5 + x(-0.5) = 16 - 2 = 14. Similarly, [x0.5 + x(-0.5)]2 = x + 2 + 1/x = 142 = 196, therefore, x + 1/x = 196 - 2 = 194. Q3. Population at 2004 = 4 million. population n years later = 8 million. So 8 = 4( 1 + 0.03)n 2 = (1.03)n log 2 = n log 1.03 n = log 2/log 1.03 = 0.301/0.0128= 23.4 So population double in 2004 + 24 = 2028.
其他解答:
Form4 maths question about logarithmic functions(3)
發問:
40.Consider 2x=5y=10z, where x,y and z are non-zero real numbers.(a) Express x,y in terms of z respectively.(b) Hence show that 1/x+1/y=1/z41.If x0.25 +x(-0.25)=4, find(a)(√x )+(1/√x)(b)x+(1/x)44.The population of a city is four million at the end of the year 204. The population grows... 顯示更多 40.Consider 2x=5y=10z, where x,y and z are non-zero real numbers. (a) Express x,y in terms of z respectively. (b) Hence show that 1/x+1/y=1/z 41.If x0.25 +x(-0.25)=4, find (a)(x )+(1/x) (b)x+(1/x) 44.The population of a city is four million at the end of the year 204. The population grows at the rate 3% per year. (a) In which year will the population be doubled? Ans: 40(a)x= z/log2 y=z/log5 41.(a)14 (b)194 44.(a)2028 唔該列式教教我點計,多謝!
最佳解答:
Q1. 2x = 10z x log 2 = z log 10 = z........(1) 5y = 10z y log 5 = z log 10 = z .........(2) So 1/x + 1/y = log 2/z + log 5/z = (log 2 + log 5)/z = (log 10)/z = 1/z. Q2. [x0.25 + x(-0.25)]2 = x0.5 + 2(x0.25)[x(-0.25)] + x(-0.5) = x0.5 + 2 + x(-0.25) = 42 = 16, therefore, x0.5 + x(-0.5) = 16 - 2 = 14. Similarly, [x0.5 + x(-0.5)]2 = x + 2 + 1/x = 142 = 196, therefore, x + 1/x = 196 - 2 = 194. Q3. Population at 2004 = 4 million. population n years later = 8 million. So 8 = 4( 1 + 0.03)n 2 = (1.03)n log 2 = n log 1.03 n = log 2/log 1.03 = 0.301/0.0128= 23.4 So population double in 2004 + 24 = 2028.
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