Rewrite formula

Question

I have the formula: $$ R(T)=R(T_0)e^{-B\left(\frac{1}{T_0} - \frac{1}{T}\right)} $$ How can I isolate $T$? I came this far: $$ \ln\left(\frac{R(T)}{R(T_0)}\right)= -B\left(\frac{1}{T_0} - \frac{1}{T}\right) \implies \frac{\ln\left(\frac{R(T)}{R(T_0)}\right)}{-B}=\frac{1}{T_0} - \frac{1}{T}$$

Answer 1

take the logarithm of both sides we obtain $\ln\left(\frac{R(T)}{R(T_0}\right)=-\frac{B}{T_0}+\frac{B}{T}$ and then $\ln\left(\frac{R(T)}{R(T_0}\right)+\frac{B}{T_0}=\frac{B}{T}$ solving for $T$ we get $T=\frac{B}{\ln\left(\frac{R(T)}{R(T_0}\right)+\frac{B}{T_0}}$

Answer 2

Now add $\frac 1T+\frac{ln(\frac{R(T)}{R(T_0)})}{B}$ to each side, which gets $\frac 1T$ alone on the left. Invert both sides

Answer 3

If the function $f :x \mapsto R(x)e^{-B/x}$ is invertible (that depends on what is exactly $R(x)$), than you can rearrange the terms from your equation as $$ f(T) = f(T_0) $$ from which $$ T = f^{-1}(f(T_0)) = T_0 $$