Looooooooooooog-ARITHMS!

Applying properties of logarithms, $\log_2 [ 2^3 4^4 8^5 \dots (2^{23})^{25} ] = \log_2 2^3 + \log_2 4^4 + \log_2 8^5 + \dots + \log_2 (2^{23})^{25} = 1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots + 23 \cdot 25.$ Recall from an advanced algebra course that $1 \cdot 3 + 2 \cdot 4 + \dots + n \cdot (n + 2) = \frac{n (n + 1) (2n + 7)}{6}.$ So, with $n = 23$ , we have $\log_2 [ 2^3 4^4 8^5 \dots ( 2^{23})^{25} ] = \frac{23 (23 + 1)[2(23) + 7]}{6} = \frac{23(24)(53)}{6} = \boxed{4876}.$

$\log _{ 2 }{ [{ 2 }^{ 3 }{ 4 }^{ 4 }{ 8 }^{ 5 }...{ ({ 2 }^{ 23 }) }^{ 25 }] } \\ The\quad pattern\quad goes\quad like\quad this.\\ { 2 }^{ 3 },\quad { 2 }^{ 8 },\quad { 2 }^{ 15 },\quad ...\quad ,\quad { (2^{ 23 }) }^{ 25 }\\ We\quad must\quad find\quad the\quad sum\quad of\quad the\quad exponents.\\ The\quad exponents\quad are\quad in\quad a\quad sequence.\\ Let\quad \{ { a }_{ n }\} :3,\quad 8,\quad 15,\quad ...,\quad 575\\ Let\quad \{ b_{ n }\} :{ a }_{ n+1 }-{ a }_{ n }\\ \{ { b }_{ n }\} :\quad 5,\quad 7,\quad 9,\quad ...\\ Finding\quad the\quad general\quad term\quad of\quad \{ { b }_{ n }\} ,\\ \{ { b }_{ n }\} :\quad 2n+3\\ Using\quad the\quad formula,\\ { a }_{ n }={ a }_{ 1 }+\sum _{ k=1 }^{ n-1 }{ { b }_{ k } } \\ { a }_{ n }=3+\sum _{ k=1 }^{ n-1 }{ (2k+3) } \\ =3+2\sum _{ k=1 }^{ n-1 }{ k } +3\sum _{ k=1 }^{ n-1 }{ 1 } \\ =3+n(n-1)+3(n-1)\\ =3+{ n }^{ 2 }-n+3n-3={ n }^{ 2 }+2n\\ So\quad the\quad general\quad term\quad of\quad { a }_{ n }\quad is\\ { a }_{ n }={ n }^{ 2 }+2n\\ The\quad last\quad term\quad of\quad { a }_{ n }\quad is\quad 575,\quad which\quad is\quad the\quad 23rd\quad term.\\ Finding\quad the\quad sum\quad of\quad the\quad exponents\quad (or\quad { a }_{ n }),\\ \sum _{ n=1 }^{ 23 }{ { n }^{ 2 } } +2\sum _{ n=1 }^{ 23 }{ n } \\ =\frac { 23(24)(47) }{ 6 } +(23)(24)\\ =4324+552=4876\\ \\ So\\ \log _{ 2 }{ [{ 2 }^{ 3 }{ 4 }^{ 4 }{ 8 }^{ 5 }...{ ({ 2 }^{ 23 }) }^{ 25 }]=\log _{ 2 }{ { 2 }^{ 4876 } } } \\ \\ \boxed { Therefore,\quad \log _{ 2 }{ [{ 2 }^{ 3 }{ 4 }^{ 4 }{ 8 }^{ 5 }...{ ({ 2 }^{ 23 }) }^{ 25 }]=\log _{ 2 }{ { 2 }^{ 4876 } } } =4876\quad } \\ \\ \\ \\ \\ \\$

First rewrite it as a summation: sum(1:23) logbase 2 of (2^(x^(x+2))) sum(1:23) logbase 2 of (2^(x(x+2)) then sum(1:23) x(x+2)logbase 2 of 2 = sum(1:23) x(x+2) = sum(1:23) x^2 + 2 sum(1:23) x = [23 (23+1) (2 (23)+1)]/6 + 2*[23(23+1)]/2 = 4324 + 552 = 4876