Reformulate a term

Question

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:

$$T= \frac{b T_0}{T_0\ln R-T_0\ln R_0+b}$$

Answer 1

1. Take the logarithm on both sides: $$ \ln R - \ln R_1 = b\left(\frac{1}{T}-\frac{1}{T_0}\right) $$

2. Divide both sides by $b\neq 0$ and add $\frac{1}{T_0}$: $$ \frac{1}{b}\left( \ln R - \ln R_1\right) + \frac{1}{T_0} = \frac{1}{T} $$

3. Take the reciprocal of both sides: $$ \frac{1}{\frac{1}{b}\left( \ln R - \ln R_1\right) + \frac{1}{T_0}} = T $$

4. Multiply both numerator and denominator of the LHS by $b$: $$ \frac{b}{\ln R - \ln R_1 + \frac{b}{T_0}} = T $$

5. Multiply both numerator and denominator of the LHS by $T_0$: $$ \frac{bT_0}{T_0\ln R - T_0\ln R_1 + b} = T $$

Answer 2

$\frac{R}{R_1}= e^{b({\frac{1}{T}-\frac{1}{T_0})}}$ 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}

Answer 3

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.