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Rodrigo de Azevedo
Rodrigo de Azevedo
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Taking the ratio of 2b) and 2a) resp. 2c) and 2b) and taking the logarithm yields \begin{eqnarray} ln\left(\frac{F-F_1}{F-F_2}\right)&=&n\cdot ln\left(\frac{a_1}{a_2}\right)\\ ln\left(\frac{F-F_2}{F-F_3}\right)&=&n\cdot ln\left(\frac{a_2}{a_3}\right) \end{eqnarray}\begin{eqnarray} \ln\left(\frac{F-F_1}{F-F_2}\right)&=&n\cdot \ln\left(\frac{a_1}{a_2}\right)\\ \ln\left(\frac{F-F_2}{F-F_3}\right)&=&n\cdot \ln\left(\frac{a_2}{a_3}\right) \end{eqnarray} Taking again the ratio yields $$ ln\left(\frac{F-F_1}{F-F_2}\right)ln\left(\frac{a_2}{a_3}\right)=ln\left(\frac{F-F_2}{F-F_3}\right)ln\left(\frac{a_1}{a_2}\right) $$$$ \ln\left(\frac{F-F_1}{F-F_2}\right)ln\left(\frac{a_2}{a_3}\right)=\ln\left(\frac{F-F_2}{F-F_3}\right)\ln\left(\frac{a_1}{a_2}\right) $$ It follows that $$ (F-F_1)^{ln(a_2)}(F-F_2)^{ln(a_3)}(F-F_3)^{ln(a_1)}=(F-F_1)^{ln(a_3)}(F-F_2)^{ln(a_1)}(F-F_3)^{ln(a_2)}, $$$$ (F-F_1)^{\ln(a_2)}(F-F_2)^{\ln(a_3)}(F-F_3)^{\ln(a_1)}=(F-F_1)^{\ln(a_3)}(F-F_2)^{\ln(a_1)}(F-F_3)^{\ln(a_2)}, $$ i.e. an equation with the only unknown $F$. If this is solved we can solve for $n$ using $$ n=\frac{ln\left(\frac{F-F_1}{F-F_2}\right)}{ln\left(\frac{a_1}{a_2}\right)}=\frac{ln(F-F_1)-ln(F-F_2)}{ln(a_1)-ln(a_2)}. $$$$ n=\frac{\ln\left(\frac{F-F_1}{F-F_2}\right)}{\ln\left(\frac{a_1}{a_2}\right)}=\frac{\ln(F-F_1)-\ln(F-F_2)}{\ln(a_1)-\ln(a_2)}. $$ Then one can solve for $C$ using one of the equations 2a),2b) resp. 2c).

Taking the ratio of 2b) and 2a) resp. 2c) and 2b) and taking the logarithm yields \begin{eqnarray} ln\left(\frac{F-F_1}{F-F_2}\right)&=&n\cdot ln\left(\frac{a_1}{a_2}\right)\\ ln\left(\frac{F-F_2}{F-F_3}\right)&=&n\cdot ln\left(\frac{a_2}{a_3}\right) \end{eqnarray} Taking again the ratio yields $$ ln\left(\frac{F-F_1}{F-F_2}\right)ln\left(\frac{a_2}{a_3}\right)=ln\left(\frac{F-F_2}{F-F_3}\right)ln\left(\frac{a_1}{a_2}\right) $$ It follows that $$ (F-F_1)^{ln(a_2)}(F-F_2)^{ln(a_3)}(F-F_3)^{ln(a_1)}=(F-F_1)^{ln(a_3)}(F-F_2)^{ln(a_1)}(F-F_3)^{ln(a_2)}, $$ i.e. an equation with the only unknown $F$. If this is solved we can solve for $n$ using $$ n=\frac{ln\left(\frac{F-F_1}{F-F_2}\right)}{ln\left(\frac{a_1}{a_2}\right)}=\frac{ln(F-F_1)-ln(F-F_2)}{ln(a_1)-ln(a_2)}. $$ Then one can solve for $C$ using one of the equations 2a),2b) resp. 2c).

Taking the ratio of 2b) and 2a) resp. 2c) and 2b) and taking the logarithm yields \begin{eqnarray} \ln\left(\frac{F-F_1}{F-F_2}\right)&=&n\cdot \ln\left(\frac{a_1}{a_2}\right)\\ \ln\left(\frac{F-F_2}{F-F_3}\right)&=&n\cdot \ln\left(\frac{a_2}{a_3}\right) \end{eqnarray} Taking again the ratio yields $$ \ln\left(\frac{F-F_1}{F-F_2}\right)ln\left(\frac{a_2}{a_3}\right)=\ln\left(\frac{F-F_2}{F-F_3}\right)\ln\left(\frac{a_1}{a_2}\right) $$ It follows that $$ (F-F_1)^{\ln(a_2)}(F-F_2)^{\ln(a_3)}(F-F_3)^{\ln(a_1)}=(F-F_1)^{\ln(a_3)}(F-F_2)^{\ln(a_1)}(F-F_3)^{\ln(a_2)}, $$ i.e. an equation with the only unknown $F$. If this is solved we can solve for $n$ using $$ n=\frac{\ln\left(\frac{F-F_1}{F-F_2}\right)}{\ln\left(\frac{a_1}{a_2}\right)}=\frac{\ln(F-F_1)-\ln(F-F_2)}{\ln(a_1)-\ln(a_2)}. $$ Then one can solve for $C$ using one of the equations 2a),2b) resp. 2c).

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user35593
user35593
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Taking the ratio of 2b) and 2a) resp. 2c) and 2b) and taking the logarithm yields \begin{eqnarray} ln\left(\frac{F-F_1}{F-F_2}\right)&=&n\cdot ln\left(\frac{a_1}{a_2}\right)\\ ln\left(\frac{F-F_2}{F-F_3}\right)&=&n\cdot ln\left(\frac{a_2}{a_3}\right) \end{eqnarray} Taking again the ratio yields $$ ln\left(\frac{F-F_1}{F-F_2}\right)ln\left(\frac{a_2}{a_3}\right)=ln\left(\frac{F-F_2}{F-F_3}\right)ln\left(\frac{a_1}{a_2}\right) $$ It follows that $$ (F-F_1)^{ln(a_2)}(F-F_2)^{ln(a_3)}(F-F_3)^{ln(a_1)}=(F-F_1)^{ln(a_3)}(F-F_2)^{ln(a_1)}(F-F_3)^{ln(a_2)}, $$ i.e. an equation with the only unknown $F$. If this is solved we can solve for $n$ using $$ n=\frac{ln\left(\frac{F-F_1}{F-F_2}\right)}{ln\left(\frac{a_1}{a_2}\right)}=\frac{ln(F-F_1)-ln(F-F_2)}{ln(a_1)-ln(a_2)}. $$ Then one can solve for $C$ using one of the equations 2a),2b) resp. 2c).