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The-vertices-of-quadrilateral-lie-on-the-graph-of-y-lnx-and-the-x-coordinates-of-these-vertices-are-consecutive-positive-integer-The-area-of-the-quadrilateral-is-ln-91-90-what-is-the-x-c




Question Number 81734 by jagoll last updated on 15/Feb/20
The vertices of quadrilateral  lie on the graph of y = lnx and   the x−coordinates of these vertices  are consecutive positive integer  . The area of the quadrilateral  is ln (((91)/(90))). what is the x−coordinate  of the leftmost vertex
$${The}\:{vertices}\:{of}\:{quadrilateral} \\ $$$${lie}\:{on}\:{the}\:{graph}\:{of}\:{y}\:=\:{lnx}\:{and}\: \\ $$$${the}\:{x}−{coordinates}\:{of}\:{these}\:{vertices} \\ $$$${are}\:{consecutive}\:{positive}\:{integer} \\ $$$$.\:{The}\:{area}\:{of}\:{the}\:{quadrilateral} \\ $$$${is}\:{ln}\:\left(\frac{\mathrm{91}}{\mathrm{90}}\right).\:{what}\:{is}\:{the}\:{x}−{coordinate} \\ $$$${of}\:{the}\:{leftmost}\:{vertex} \\ $$
Commented by john santu last updated on 15/Feb/20
12?
$$\mathrm{12}?\: \\ $$
Commented by jagoll last updated on 15/Feb/20
solution pls
$${solution}\:{pls} \\ $$
Commented by mr W last updated on 15/Feb/20
A=−ln n+ln (n+1)+ln (n+2)−ln (n+3)  A=ln (((n+1)(n+2))/(n(n+3)))=ln ((91)/(90))  ⇒n^2 +3n−180=0  ⇒(n+15)(n−12)=0  ⇒n=12
$${A}=−\mathrm{ln}\:{n}+\mathrm{ln}\:\left({n}+\mathrm{1}\right)+\mathrm{ln}\:\left({n}+\mathrm{2}\right)−\mathrm{ln}\:\left({n}+\mathrm{3}\right) \\ $$$${A}=\mathrm{ln}\:\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{{n}\left({n}+\mathrm{3}\right)}=\mathrm{ln}\:\frac{\mathrm{91}}{\mathrm{90}} \\ $$$$\Rightarrow{n}^{\mathrm{2}} +\mathrm{3}{n}−\mathrm{180}=\mathrm{0} \\ $$$$\Rightarrow\left({n}+\mathrm{15}\right)\left({n}−\mathrm{12}\right)=\mathrm{0} \\ $$$$\Rightarrow{n}=\mathrm{12} \\ $$
Commented by jagoll last updated on 15/Feb/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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