Easy exponential equation but…

Question

$$\left(\sqrt{5+2\sqrt6}\right)^x-\left(\sqrt{5-2\sqrt6}\right)^x=96$$ I was trying to use that $$(5+2\sqrt6)(5-2\sqrt6)=1$$ but I end up with $$t^2-98t+1=0$$ which has some ugly solutions and I think I should have some nice ones there. Thx

Answer 1

It should be $t^2-96t-1=0,$ which gives $$t=48+\sqrt{2305}$$ and $$x=\log_{\sqrt{5+2\sqrt{6}}}(48+\sqrt{2305})$$ or $$x=\log_{\sqrt2+\sqrt3}(48+\sqrt{2305})$$

Answer 2

$$\left(\sqrt{5+2\sqrt{6}}\right)^x - \left(\sqrt{5-2\sqrt{6}}\right)^x = 96$$ Take natural log of both sides $$\ln \left[ \left(\sqrt{5+2\sqrt{6}}\right)^x - \left(\sqrt{5-2\sqrt{6}}\right)^x \right]= \ln96$$ Use the $\ln(a+b) = \ln(a) + \ln(1+b/a)$ property $$\ln \left(\sqrt{5+2\sqrt{6}}\right)^x + \ln \left(1-\dfrac{\left(\sqrt{5-2\sqrt{6}}\right)^x}{\left(\sqrt{5+2\sqrt{6}}\right)^x}\right)=\ln96$$ Take out the $x$ from the log $$x \left[\ln\left(\sqrt{5+2\sqrt{6}}\right)\right] + x\left[\ln\left(1-\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}}}\right)\right] =\ln96$$ Factor the $x$ and isolate it $$x=\dfrac {\ln96} {\ln\left(\sqrt{5+2\sqrt{6}}\right) + \ln\left(1-\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}}}\right)}$$ The rest is just computation