How did we got this?
We had to find $T$.
From:
$$\frac{R}{R_1}= e^{b(\frac{1}{T}-\frac{1}{T_0})}$$
This:
$$T= \frac{b T_0}{T_0\ln R-T_0\ln R_0+b}$$
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What have you attempted? Specifically, taking the logarithm on both sides -- did you try that? – Clement C. Jul 05 '16 at 14:27
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Of course I did, but after doing that I do not come to the formula. – Jul 05 '16 at 14:30
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See my answer, then. – Clement C. Jul 05 '16 at 14:34
3 Answers
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Take the logarithm on both sides: $$ \ln R - \ln R_1 = b\left(\frac{1}{T}-\frac{1}{T_0}\right) $$
Divide both sides by $b\neq 0$ and add $\frac{1}{T_0}$: $$ \frac{1}{b}\left( \ln R - \ln R_1\right) + \frac{1}{T_0} = \frac{1}{T} $$
Take the reciprocal of both sides: $$ \frac{1}{\frac{1}{b}\left( \ln R - \ln R_1\right) + \frac{1}{T_0}} = T $$
Multiply both numerator and denominator of the LHS by $b$: $$ \frac{b}{\ln R - \ln R_1 + \frac{b}{T_0}} = T $$
Multiply both numerator and denominator of the LHS by $T_0$: $$ \frac{bT_0}{T_0\ln R - T_0\ln R_1 + b} = T $$
Clement C.
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Incidentally, this is under the (implicit) assumption $b\neq 0$, and that the $R_0$ written in your question is actually a typo for $R_1$. – Clement C. Jul 05 '16 at 14:34
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$\frac{R}{R_1}= e^{b({\frac{1}{T}-\frac{1}{T_0})}}$ Take logarithm to the base $e$ on both sides. ln(R/R1)= b{(1/T)-(1/T0)} {1/b}{ln R - lnR1}= {(1/T)-(1/T0)} (As ln (a/b)= ln a - ln b) 1/T = {T0(ln R - ln R1) - b}/{T0*b} Reciprocal on both sides T={T0*b}/{T0(ln R - ln R1) - b}
sai saandeep
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Start of by taking the ln on both sides. Note: that after you take the ln on both side ln(e^x)=x: This is what you are missing.
$Ln(\frac{R}{R1})$=$b*(\frac{1}{T}-\frac{1}{T_0})$
then use properties of logs and algebra to simplify to whatever you need.
