Maths(log)
2007-09-05 8:24 pm
A line x = j intersects the graph of y = log5 x and the graph of y = log5 (x+4) at points A and B. The distance AB is 0.5. Given that j = c + square root d, where c and d are integers, what is c +d?
回答 (2)
2007-09-06 12:34 am
✔ 最佳答案
As follows:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Sep07/Crazyeqn1.jpg
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Sep07/Crazyeqn2.jpg
參考: My Maths knowledge
2007-09-06 12:13 am
x=j --- (1)
y=log5j --- (2)
y=log5(j+4) --- (3)
Using (1) & (2), Coord. A=(j , log5j)
Using (1) & (3), Coord. B=(j, log5(j+4)
AB=0.5 = log5(j+4) - log5 = log [(j+4)/j]
i.e. log (sqrt 10) = log [(j+4)/j]
finally, u can simpify the eqt. as 9 (squ j) - 8j - 16 =0 and
j = (4/9) + sqrt (160/81) = c + sqrt d
that is, c = 4/9 & d = 160/81
Therefore, c+d = 196/81
Remark: c, d are not iterger
y=log5j --- (2)
y=log5(j+4) --- (3)
Using (1) & (2), Coord. A=(j , log5j)
Using (1) & (3), Coord. B=(j, log5(j+4)
AB=0.5 = log5(j+4) - log5 = log [(j+4)/j]
i.e. log (sqrt 10) = log [(j+4)/j]
finally, u can simpify the eqt. as 9 (squ j) - 8j - 16 =0 and
j = (4/9) + sqrt (160/81) = c + sqrt d
that is, c = 4/9 & d = 160/81
Therefore, c+d = 196/81
Remark: c, d are not iterger
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