+118608 It isn't. It is just a huge number.
Too huge for the calculator to deal with. :)
+118608 I decided to practice my skills and see if I can get an approximation of \(2^{2015}\)
\(let \;\;x=2^{2015}\\ log_2x=2015\\ \frac{log_{10}x}{log_{10}2}=2015\\ log_{10}x=2015 \times log_{10}2\\ log_{10}x\approx 606.5754413\qquad\mbox{From the calculator}\\ x\approx 10^{ 606.5754413}\\ x\approx 10^{ 0.5754413}\times10^{ 606}\\ x\approx 3.762194983\times10^{ 606}\\~\\ 2^{2015}\approx 3.762194983\times10^{ 606}\\\)
Oh I made it slightly longer than necessary.
I could just have said
\(x=2^{2015}\\ logx=log2^{2015}\qquad \mbox{both base 10}\\ logx=2015*log2\\ etc\)
