Simple log math

Question

So I have the equation $$-2 \frac{\mu_{0} Il}{2\pi}\ln \Biggr( \frac{2h}{\sqrt{(2h)^2 +(2d)^2}} \Biggr)$$ and this should evaluate to $$ \frac{\mu_{0}Il}{2 \pi} \ln \Biggr(1 + \frac{d^2}{h^2} \Biggr) $$ So I think that $\frac{\mu_{0} Il}{2\pi}$ gets irrelevant. I thought of getting $2h$ down to the denominator like $$ln \Biggr( \frac{1}{\frac{1}{2h} \sqrt{(2h)^2 + (2d)^2}} \Biggr)$$ and then multiply it into the square root, so it should evaluate to $$\ln \Biggr( \frac{1}{\sqrt{1+ \frac{d^2}{h^2}}} \Biggr)$$ But now how do I get from $$-2\ln \Biggr( \frac{1}{\sqrt{1+ \frac{d^2}{h^2}}} \Biggr)$$ to $$\ln \Biggr(1+ \frac{d^2}{h^2} \Biggr) $$

Answer 1

The rules you need are that $\ln(x^n)=n\ln (x)$ and $-\ln(\frac 1x)=\ln(x)$, here used in reverse. The second is the same as the first with $n=-1$. You are correct the leading $\frac {\mu_0Il}{2\pi}$ doenn't matter as it just carries through. Then $$-2\ln \Biggr( \frac{2h}{\sqrt{(2h)^2 +(2d)^2}} \Biggr)=2\ln\left( \frac{\sqrt{(2h)^2 +(2d)^2}}{2h} \right)\\=\ln\left( \frac{(2h)^2 +(2d)^2}{(2h)^2} \right)=\ln\left(1+\frac{d^2}{h^2}\right)$$

Answer 2

we have $$\ln\left(\frac{2h}{\sqrt{(2h)^2+(2d)^2}}\right)=\ln(2h)-\ln(\sqrt{(2h)^2+(2d)^2})=\ln(2h)-\ln\left(2h\sqrt{1+\left(\frac{d}{h}\right)^2}\right)=\ln(2h)-\ln(2h)-\ln\left(\sqrt{1+\left(\frac{d}{h}\right)^2}\right)$$