Why does $\ln(2+\sqrt3)={1\over2}\ln(7+4\sqrt3)$?

Question

I was solving the definite integral $\int_{\sqrt7}^{2\sqrt7}{1\over\sqrt {x^2-7}}dx$, and came out with the intermediate step $\int_{\sqrt7}^{2\sqrt7}\sec\theta\ d\theta$, which led me to finish off with $$\ln \left|\sec\theta +\tan\theta \right|\bigg|_{arc\sec(1)}^{arc\sec(2)}$$

$$=\ln\left|x+\sqrt{x^2-7}\right|\bigg|_{\sqrt7}^{2\sqrt7}-\require{cancel} \cancel{\ln\sqrt7}$$

$$=\ln\left|{2\sqrt7+\sqrt{21}}\over\sqrt7\right|$$

$$=\ln\left|{2+\sqrt{3}}\right|$$

However, my book, after the step of $\int_{\sqrt7}^{2\sqrt7}\sec\theta\ d\theta$, seems to go in a completely different direction and comes out with $$={1\over2}\ln\left|{7+4\sqrt{3}}\right|$$

Upon checking them, I see that they are equal, and my question is, is there any property of logarithms or algebraic reason with which I can recognize them as equal, without solving the integral $\int_{\sqrt7}^{2\sqrt7}\sec\theta\ d\theta$ in a different way?

Answer 1

Observe that: $$ \ln(2+\sqrt3)=\ln(((2+\sqrt3)^2)^\frac{1}{2})=\frac{1}{2}\ln(4 +3+4\sqrt3) = {1\over2}\ln(7+4\sqrt3)$$

Here we used $\ln(a^b)=b \ln (a)$.

Answer 2

Just square $2 + \sqrt{3}$ and you will see it. ;-)

Answer 3

You have to verify whether $-2\ln(2-\sqrt3)=\ln(7+4\sqrt3)$ is true, that is $$(2-\sqrt3)^{-2}=(7+4\sqrt3)$$

or $$(2-\sqrt3)^2(7+4\sqrt3)=1$$

or $$(7-4\sqrt3)(7+4\sqrt3)=1,$$

which is obviously true, since $49-48 =1$.

Answer 4

Hint $:$ $7+4 \sqrt 3 = (2+\sqrt 3)^2$.