## Finding normal to ln curve

I was wondering if anyone could help me with a question I was trying to work through today in our A-level math class.

1. The problem statement, all variables and given/known data

a. Sketch the graph of y= ln(3x)
b. The normal to the curve at Point Q, with x coordinate q, passes through the origin. Find the equation of the normal and the value of q

2. Relevant equations

I don’t know which to use

3. The attempt at a solution

Part a was not too difficult and I was able to do that

However I have no idea how to tackle part b. I have only been able to express the gradient of the normal as -q, but cannot solve it. I think [but am not sure] that I can get to the equation

ln3q + q^2 = 0

but I cannot solve this.

I would be very grateful for some help.

Thanks

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 Since when the normal and the curve y=ln(3x) intersect, the x coordinate is q, the the y coordinate is thus ln3q. And since the normal passes through (0,0), the gradient of the normal can also be expressed as (ln3q)/q I believe also that the gradient of the normal can be expressed as -q, thus (ln3q) / q= -q Rearranging gives ln3q + q^2 = 0 However I don't know any method for solving such an equation

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## Finding normal to ln curve

 Quote by Doc G Since when the normal and the curve y=ln(3x) intersect, the x coordinate is q, the the y coordinate is thus ln3q.
Good
 Quote by Doc G And since the normal passes through (0,0), the gradient of the normal can also be expressed as (ln3q)/q
Would it not be better to express the gradient as;

$$m=\frac{dy}{dx}\ln\left(3x\right) = \frac{1}{x}$$

$$\left.\frac{dy}{dx}\right|_{\left(q,\ln(q)\right)} = \frac{1}{q}$$

 Recognitions: Gold Member Science Advisor Staff Emeritus Almost. That is the slope of the tangent line. The slope of the normal line is -3q
 Thanks for the replies. Okay we haven't really covered much differentiation of ln graphs yet, but using what hootenanny/HallsofIvy posted, should the equation therefore be ln(3q)/q = -3q giving: ln(3q) + 3(q^2) = 0 is this right? and how could i solve that? thanks again

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 Quote by Hootenanny Good Would it not be better to express the gradient as; $$m=\frac{dy}{dx}\ln\left(3x\right) = \frac{1}{3x}$$
 Thanks, should I now attempt to find a close solution by iteration [we've only just covered the basics of which in class]? And would the correct formula for which be: $$x_{n+1} = x_{n} - \frac{\ln3x_{n} + x^2_{n}}{1/x_{n} + 2x}$$ ? i tried putting in the equation ln3q + q^2 = 0 into Newton's Method. Is this okay? - i dont really have a good understanding of this area of maths yet..