1. ## Approximate values

The table below gives values of T and L which are related by the law $T = kL^n$ where k and n are constants.

T - __1____1.2___1.4___1.6___1.8___2
L - __2.4___3.5__ 4.8__ 6.3___7.9___9.8

Prove that this law is as stated. Determine the approximate values of k and n, and hence state the law.

2. Originally Posted by Dean
The table below gives values of T and L which are related by the law $T = kL^n$ where k and n are constants.

T - __1____1.2___1.4___1.6___1.8___2
L - __2.4___3.5__ 4.8__ 6.3___7.9___9.8

Prove that this law is as stated. Determine the approximate values of k and n, and hence state the law.
Consider
$T = kL^n$

Take the log (or natural log, or log to your favorite base, whatever) of both sides:
$log(T) = log(kL^n) = log(L^n) + log(k)$

$log(T) = n \cdot log(L) + log(k)$

What does this do for you? Well the equation is now in the form
$y = m \cdot x + b$
where $log(T) = y, \ n = m, \ log(L) = x, \ log(k) = b$

So what you do is a linear regression with the log(L) values being your x values and the log(T) values being the y values. Then the slope m of the regression will be n, and the intercept b will be the log(k).

Edit: For reference I get that $n \approx 0.493468$ and $log(k) \approx -0.188847 \implies k \approx 0.827913$ with an $r^2 \approx 0.999885$, so the fit for $T = (0.827913)L^{0.493468}$ is excellent. This is not quite the same as Soroban's answer (which is easier to get) but is probably more accurate. (Sorry Soroban, I'm not trying to pick on your answer! )

-Dan

3. Hello, Dean!

The table below gives values of $T$ and $L$
which are related by the law $T \:= \:kL^n$ where $k$ and $n$ are constants.

$\begin{array}{cccccccc}T & | & 1 & 1.2 & 1.4 & 1.6 & 1.8 & 2 \\ \hline
L & | & 2.4 & 3.5 & 4.8 & 6.3 & 7.9 & 9.8
\end{array}$

Prove that this law is as stated.
Determine the approximate values of $k$ and $n$, and hence state the law.

Use any two pairs of values: . $(L,\,T) \:=\:(2.4,\,1),\;(3,5,\,1.2)$

Substitute into the law:

. $\begin{array}{ccccc}(2.4,\,1): & k\cdot2.4^n & = & 1 & {\color{blue}[1]} \\
(3,5,\,1.2): & k\cdot3.5^n & = & 1.2 & {\color{blue}[2]} \end{array}$

Divide ${\color{blue}[2]}$ by ${\color{blue}[1]}$: . $\frac{k\cdot3.5^n}{k\cdot2.4^n} \:=\:\frac{1.2}{1}\quad\Rightarrow\quad \left(\frac{3.5}{2.4}\right)^n \:=\:1.2$

Take logs: . $\ln\left(\frac{3.5}{2.4}\right)^n \:=\:\ln(1.2)\quad\Rightarrow\quad n\cdot\ln\left(\frac{3.5}{2.4}\right) \:=\:\ln(1.2)$

. . Hence: . $n \;=\;\frac{\ln(1.2)}{\ln\left(\frac{3.5}{2.4}\righ t)} \;=\;0.483234414$

Substitute into ${\color{blue}[1]}$: . $k\cdot2.4^{0.483234414} \;=\;1\quad\Rightarrow\quad k \:=\:0.655041542$

Therefore: . $\begin{array}{ccc}n & \approx & 0.483 \\ k & \approx & 0.655\end{array}$ . and the law is: . $T \;=\;0.655L^{0.483}$