Re: Combining two equations

$\displaystyle L=k_1 \cdot \frac{M^6}{R^7} ~\text {and}~L=k_2 \cdot \frac{M^{5.5}}{R^{0.5}}$

$\displaystyle k_1 \cdot \frac{M^6}{R^7}=k_2 \cdot \frac{M^{5.5}}{R^{0.5}} \Rightarrow k_1 \cdot M^{0.5}=k_2 \cdot R^{6.5} \Rightarrow M=\left(\frac{k_2}{k_1}\right)^2 \cdot R^{13}$

$\displaystyle L=k_1 \cdot \left(\frac{k_2}{k_1}\right)^{12} \cdot \frac{R^{78}}{R^7} \Rightarrow L=k_3 \cdot R^{71}$

Re: Combining two equations

Hello, ohdearthisisheadrot!

Quote:

I have two equations as follows:

. . L is proportional to (M^6/R^7)

. . L is proportional to M^5.5 R^-0.5

I need to end up with a relationship which has L proportional to R so that M is eliminated.

To eliminate $\displaystyle M$, solve both equations for $\displaystyle M$ and equate.

We have: .$\displaystyle \begin{Bmatrix}L &=& a\dfrac{M^6}{R^7} \\ \\[-3mm] L &=& b\dfrac{M^{\frac{11}{2}}}{R^{\frac{1}{2}}} \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}M^6 &=& \dfrac{LR^7}{a} & [1] \\ \\[-3mm] M^{\frac{11}{2}} &=& \dfrac{LR^{\frac{1}{2}}}{b} & [2] \end{Bmatrix} $

$\displaystyle \begin{array}{cccccccccccc}\text{Raise [1] to the power }\frac{11}{2}\!: & \left(M^6\right)^{\frac{11}{2}} &=& \left(\dfrac{LR^7}{a}\right)^{\frac{11}{2}} & \Rightarrow & M^{33} &=& \dfrac{L^{\frac{11}{2}}R^{\frac{77}{2}}}{a^{\frac{ 11}{2}}} & [3]\\ \\ \text{Raise [2] to the power }6\!: & \left(M^{\frac{11}{2}}\right)^6 &=& \left(\dfrac{LR^{\frac{1}{2}}}{b}\right)^6 & \Rightarrow & M^{33} &=& \dfrac{L^6R^3}{b^6}& [4] \end{array}$

Equate [3] and [4]: .$\displaystyle \dfrac{L^{\frac{11}{2}}R^{\frac{77}{2}}}{a^{\frac{ 11}{2}}} \;=\;\dfrac{L^6R^3}{b^6} \quad\Rightarrow\quad b^6L^{\frac{11}{2}}R^{\frac{77}{2}} \;=\;a^{\frac{11}{2}} L^6 R^3 $

Divide by $\displaystyle L^{\frac{11}{2}}R^3\!:\;\;b^6R^{\frac{71}{2}} \;=\;a^{\frac{11}{2}} L^{\frac{1}{2}} \quad\Rightarrow\quad L^{\frac{1}{2}} \;=\;\dfrac{b^6R^{\frac{71}{2}}}{a^{\frac{11}{2}}} $

Square both sides: . $\displaystyle L \;=\;\frac{b^{12}}{a^{11}}R^{71}$

Re: Combining two equations

Sorry my mistake on the initial question:

L = M^5.5*R^-0.5 not L = M^5.5/R^-0.5

does this make a difference?

thanks for your help!