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i'm going through some homework, but there is one thing i don't understand. Our task is to explain the following calculation:

Given: $h(x) = \ln(x^4)$

I.
$ \begin{align} h(x) = t(x) &\Rightarrow \ln(x^4) = mx = 4 \\ &\Rightarrow x^4 = e^4 \\ &\Rightarrow x = e \quad \textrm{or}\quad x = -e \end{align} $

I am confused about the line: $\Rightarrow \ln(x^4) = mx = 4$.

II.

\begin{align} h'(x) = m &\Rightarrow 4/x = m \\ &\Rightarrow x = 4/m \\ &\Rightarrow t(x) = \frac{4x}{e} \end{align}

I just can't figure out what hes doing here :/

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    Hi testiii, I converted your math to LaTeX. Can you make sure I copied the problem correctly?2011-10-24

1 Answers 1

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This looks like the determination of the point at which the tangent to the graph of $h$ passes through the origin.

Hence one is looking for $x_0$ and $m$ such that $h(x_0)=mx_0$ and h'(x_0)=m. Then the line of equation $y=t(x)$ with $t(x)=mx$ is the tangent to the graph of $h$ at the point $x_0$.

Since $h(x)=4\log x$ and h'(x)=4/x, this reads as $4\log x_0=mx_0$ and $4/x_0=m$. Hence $x_0m=4$, $\log x_0=1$, $x_0=\mathrm e$, $m=4/\mathrm e$, and the equation of the tangent is $y=4x/\mathrm e$.

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    Thank you, you explained it much better than my teacher! :)2011-10-24