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Stupidly hard!

The curve with equation y=ln3x crosses the x -axis at point P(p,0).
a) Sketch the graph of y=ln3x, showing the exact value of p.

I have done this part - it crosses the x-axis at (1/3, 0).

b) The normal to the curve at the point Q, with x-coordiate q, passes through the origin. Show that x=q is a solution to the equation x^2+ln3x =0

Help with part (b) would be great.....please explain it like you're talking to a four yr old tho!
Cheers

Reply 1

KAISER_MOLE

Righto. for the normal at q to pass through the origin, we can write the equation of the normal as y=m_normalx

the gradient of the normal =-1/m(tangent)

m(tangent) = d(ln3x)/dx = 1/x = 1/q for the point q

=> equation of normal is

y=-qx

this meets y=ln3x at the point with x coord q

=> -q² = ln3q

=> ln3q + q² = 0

so q is a root of the equation