calculation - Verifying an identity of an equation for Black Scholes formula

I just started working on the Black Scholes formula with help of the book Financial option valuation by Higham. Apparently you are possible to derive the following function:

$\log(\frac{SN'(d_1)}{e^{-r(T-t)}EN'(d_2)}) = 0$

From the Black scholes formula:
$C(S,t)=SN(d_1)-Ee^{-r(T-t)}N(d_2)$

I've been puzzling arround but I'm stuck. This is where I came so far, do you know where I'm going wrong?

$\log(\frac{SN'(d_1)}{e^{-r(T-t)}EN'(d_2)}) = \log(SN'(d_1))-\log(e^{-r(T-t)}EN'(d_2))=0$

Answer

I am trying to fill in what Richard left for the second part: \begin{align*} \exp(-r(T-t))E\, N'(d_2) &= \frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\, \exp\left(-\frac{1}{2}d_2^2\right) \\ &=\frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\, \exp\left(-\frac{1}{2}\big(d_1-\sigma\sqrt{T-t}\,\big)^2\right) \\ &=\frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\\ &\qquad\qquad \exp\left(-\frac{1}{2} d_1^2 -\frac{1}{2}\sigma^2 (T-t) + d_1 \sigma\sqrt{T-t}\right)\\ &=\frac{1}{\sqrt{2\pi}}\exp(-r(T-t))E\\ &\qquad\qquad \exp\left(-\frac{1}{2} d_1^2 +\ln(S/E) + r(T-t)\right)\\ &=\frac{1}{\sqrt{2\pi}} S \, \exp\left(-\frac{1}{2}d_1^2\right)\\ &= SN'(d_1). \end{align*} That is, \begin{align*} \ln\frac{SN'(d_1)}{\exp(-r(T-t))E\, N'(d_2)} = 0. \end{align*}