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For $k=2,3,4..$ and for $x>0$ we have the functions $f_k(x) = ^k\log(x)$. The line $x=e$ intersects the x-axis in point $E$ and the graph of $f_k$ in point $P_k$. We draw the tangents to the functions $f_2$, $f_3$, $f_4$ In the points $P_2$, $P_3$ and $P_4$. Prove that these tangents go through the origin.

So in a previous exercise I encountered that if you have a tangent to a curve $f$ in a point $P$, which goes through the origin, the x-coordinate of P will satisfy $$f'(x) = \dfrac{f(x)}{x}$$

So I used this method to try solving the problem. I got: $$\dfrac{x\ln(x) - x}{\ln(k)} = \dfrac{^k\log(x)}{x}$$

My thought behind this was that if this yields $x=e$, I would have solved the problem. But it doesn't, so I'm in need for assistance. Could anyone provide me with a reason why this is incorrect, and a push in the right direction?

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    $\begingroup$ What does $^k\log(x)$ denote (how is it computed for particular $k,x$ -- is it an iterate of $\log$ or what? $\endgroup$ – coffeemath Jan 29 '14 at 21:59
  • $\begingroup$ @coffeemath It is a group of functions, the k is just any positive integer except 1, up to infinity, I'm guessing from the ellipsis. $\endgroup$ – Phaptitude Jan 29 '14 at 22:01
  • $\begingroup$ Oh, the $k$ is just the base. Is that your confusion? That's how we write bases here. $\endgroup$ – Phaptitude Jan 29 '14 at 22:09
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So $f_k(x)=\dfrac{\ln(x)}{\ln(k)}$ using change of base formula for $\log_k(x)$, and then $f'(x)=\dfrac{1}{x \ln(k)}.$ On the vertical line $x=e$ this is $\dfrac{1}{e \ln (k)}.$ This is the same as what you have set up as the value of $\dfrac{f_k(x)}{x}$ when $x=e$, which would be $\dfrac{\ln(e)/\ln(k)}{e}$ because $\ln e=1.$

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  • $\begingroup$ Ahh, so my derivative was incorrect, thanks. I used the integral instead of the derivative. I'm such a discombobulated nincompoop. $\endgroup$ – Phaptitude Jan 29 '14 at 22:23
  • $\begingroup$ Yes I thought that derivative looked odd, but familiar (the antiderivative). It also seems in your try you didn't plug in $x=e$ to the two sides. $\endgroup$ – coffeemath Jan 29 '14 at 22:43
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What you said you need to prove is correct and I hope you have basic intuition for that besides "in previous exercise I encountered..."

Now, $f_k(x)=^k log(x)=\frac{\ln(x)}{\ln(k)}$ and $f'(x)=\frac{1}{x\ln(k)}$, so check your derivative on the LHS.

As for why you need $f'(x)=\frac{f(x)}{x}$, I hope this might help: enter image description here

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  • $\begingroup$ Yeah, I proved it in that exercise. Thanks. $\endgroup$ – Phaptitude Jan 29 '14 at 23:17
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    $\begingroup$ I added some pictures for the property to illustrate what I meant by intuition. :) $\endgroup$ – Vadim Jan 30 '14 at 3:53

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