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Logs question

The fquestion is a curve C has equation Y = x - 2 - 2 log (to the base 10) x

The vertical lines x = 6 and x =7 meet at the curve at points R and S

the first part of the question is find the Y coordinate of R which i got as 4 - Log (base 10) 36. The next part is the area of the trapezium bounded be the lines X = 6 and X = 7 and the x axis is X square units, show that A = p/2 - log to the base 10 Q, I dont get this, I found the Y coordinate of S which i got as 5 - Log 49 (base 10)
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Zacken
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Original post by SunDun111
The fquestion is a curve C has equation Y = x - 2 - 2 log (to the base 10) x

The vertical lines x = 6 and x =7 meet at the curve at points R and S

the first part of the question is find the Y coordinate of R which i got as 4 - Log (base 10) 36. The next part is the area of the trapezium bounded be the lines X = 6 and X = 7 and the x axis is X square units, show that A = p/2 - log to the base 10 Q, I dont get this, I found the Y coordinate of S which i got as 5 - Log 49 (base 10)


You'll need to post the question.

You'd be best served by sketching the trapezium and remembering GCSE formula for the area of a trapezium. What are the lengths of the parallel sides (look at your sketch). What is the height/width of the trapezium (look at your sketch). What is the formula for the area? Plug in the relevant lengths.

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