# The areabounded by the axes of reference and the normal to y=logxat the point (1, 0) is

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Curve y = log x we take the base as e (not decimal)

slope of tangent = dy/dx = 1 / x

at (1, 0) slope = 1

so slope of normal at (1,0) = -1 as the product of slopes is -1..

equation of normal is y = - x + c

as it passes throug h (1,0), c = 1

so normal is : y = 1 - x

y-intercept of this normal is : 1

Area of the traingle formed by the normal to the curve, and the axes is

1/2 * 1 * 1 = 1/2

slope of tangent = dy/dx = 1 / x

at (1, 0) slope = 1

so slope of normal at (1,0) = -1 as the product of slopes is -1..

equation of normal is y = - x + c

as it passes throug h (1,0), c = 1

so normal is : y = 1 - x

y-intercept of this normal is : 1

Area of the traingle formed by the normal to the curve, and the axes is

1/2 * 1 * 1 = 1/2