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$$\frac{3\log_x y}{(\log_y x)^4}=\frac{32}{81}$$

$$y=x^{\frac23}$$

I cannot seem to find any way to "show that the equation of the normal is $y = -3x + 28$ " without using implicit differentiation.

This is from my friend's practice papers. He is studying for the O levels, so only basic differentiation is tested. Is there any way to solve this with the method limitations?

Ong Hai Xiang
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    You could at least have the image in the proper orientation, or use the MathJax formatting this site supports (preferred): https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Sean Roberson Aug 09 '17 at 17:44

1 Answers1

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Note that $$ \frac{\log_xy}{(\log_yx)^4}=\frac{\frac{\ln y}{\ln x}}{\frac{\ln^4 x}{\ln^4 y}}=\frac{\ln^5y}{\ln^5x},$$ so $$ \frac{3\log_xy}{(\log_yx)^4}=\frac{32}{81}$$ boils down to $\frac{\ln y}{\ln x}=\frac 23$, or $$\tag1 y=x^{\frac23}.$$

The derivative of $(1)$ is $y'=\frac23x^{-\frac13}$, so that the normal through the point $(x_0,x_0^ {\frac23})$ is given by $y=mx+b$, where we must take $m=-\frac1{y'(x_0)}=-\frac 32x_0^{\frac13}$, and determine $b$ such that it $x_0^ {\frac23} = -\frac 32x_0^{\frac13}\cdot x_0+b$, i.e., $b=x_0^ {\frac23} +\frac 32x_0^{\frac43}$.

Hagen von Eitzen
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  • Hi there Hagen, thanks for replying to my post. I added the equation of the normal as $y = -3x + 28$.

    Using your answer as a guide, i would get $y = (-\frac 32 x_0 ^\frac 13)x + \frac 52 x_0 ^\frac 23 $. However, this leaves me with unknown value $x_0$ which I feel is crucial to prove the exact normal equation. Do you have any method in mind?

     – Ong Hai Xiang Aug 09 '17 at 18:02
  • @OngHaiXiang Of course the normal to the curve depends on the point of the curve you consider the normal at. It seems that $y=-3x+28$ is one of the normals, namely the one passing through the specific point $(8,4)$ – Hagen von Eitzen Aug 09 '17 at 18:20