I'm trying to solve $E$ from a relation $$j_1\cdot10^{((E-E_1)/b_1)}=j_2\cdot10^{((E-E_2)/b_2}$$ My attempt was to take $\log_{10}$ from both sides, leading to $$\log_{10}(j_1)+\frac{E-E_1}{b1}=\log_{10}(j_2)+\frac{E-E_2}{b_2}$$ $$\frac{E-E_1}{b_1}-\frac{E-E_2}{b_2}=\log_{10}(j_2)-\log_{10}(j_1)$$ $$E\cdot\left(\frac{E_2}{b_2}-\frac{E_1}{b_1}\right)=\log_{10}\left(\frac{j_2}{j_1}\right)$$ $$E=\log_{10}\left(\frac{j_2}{j_1}\right)\cdot \left(\frac{E_2}{b_2}-\frac{E_1}{b_1}\right)^{-1}$$
However, this is apparently wrong, since as this is a physical problem (Tafel equation for calculating corrosion potential) and $E$ should have units of volts, but now it is dimensionless, because also $b$ has units of volts. Also, when comparing the numerical solution between this and one given by Wolfram Alpha, the values are not the same.
So my question is, what am I doing wrong here?
$$\frac{E-E_1}{b_1} - \frac{E-E_2}{b_2} =E\left(\frac{1}{b_1}-\frac{1}{b_2}\right)- \frac{E_1}{b_1}+\frac{E_2}{b_2} \neq E\left(\frac{E_2}{b_2} - \frac{E_1}{b_1}\right)$$