An algebra problem by Albert Go

$\log _{ 4 }{ \{ { [\frac { \log _{ 4 }{ \frac { 256 }{ 64 } } }{ \sin { 32 } \cos { 13 } +\cos { 32 } \sin { 13 } } ] }^{ 2\log _{ 2 }{ 4 } } } \} \quad -1\quad$

$=\log _{ 4 }{ \{ { [\frac { \log _{ 4 }{ \frac { 256 }{ 64 } } }{ \sin { 45 } } ] }^{ 2\log _{ 2 }{ 4 } } } \} \quad -1$

$=\log _{ 4 }{ \{ { [\frac { \log _{ 4 }{ \frac { 256 }{64 } } }{ 1/\sqrt { 2 } } ] }^{ 2\log _{ 2 }{ 4 } } } \} \quad -1$

$=\log _{ 4 }{ \{ { { [\sqrt { 2 } \times \log _{ 4 }{ \frac { 256 }{64 } } }{ }] }^{ 2\log _{ 2 }{ 4 } } } \} \quad -1$

$=\log _{ 4 }{ \{ { { [\sqrt { 2 } \times \log _{ 4 }{ 4 } }] }^{ 2\log _{ 2 }{ 4 } } } \} \quad -1$

$=\log _{ 4 }{ \{ { { [\sqrt { 2 } \times 1 }] }^{ 2\log _{ 2 }{ 4 } } } \} \quad -1$

$=\log _{ 4 }{ \{ { { [\sqrt { 2 } }] }^{ 2\log _{ 2 }{ 4 } } } \} \quad -1$

$=\log _{ 4 }{ \{ { { [\sqrt { 2 } }] }^{ 4 } } \} \quad -1$

$=4\log _{ 4 }{ \{ \sqrt { 2 } } \} \quad -1$

$=2\log _{ 2 }{ \{ \sqrt { 2 } } \} \quad -1$

$=\log _{ \sqrt { 2 } }{ \{ \sqrt { 2 } } \} \quad -1$

$=1\quad -1$

$= 0$

$\begin{aligned} x & = \log_4\left(\left(\frac {\log_4 \left(\frac {256}{64}\right)}{\blue{\sin 32^\circ \cos 13^\circ + \cos 32^\circ \sin 13^\circ}}\right)^{2\log_2 4}\right) - 1 & \small \blue{\text{Note that }\sin (A+B) = \sin A \cos B + \cos A \sin B} \\ & = \log_4\left(\left(\frac {\log_4 \left(4\right)}{\blue{\sin 45^\circ}}\right)^{2\times 2}\right) - 1 \\ & = \log_4\left(\left(\frac 1{\frac 1{\sqrt 2}}\right)^4\right) - 1 \\ & = \log_4\left(\left(2^\frac 12 \right)^4\right) - 1 \\ & = \log_4\left(2^2\right) - 1 \\ & = \log_4\left(4\right) - 1 \\ & = 1 - 1 = \boxed{0} \end{aligned}$

First work inside the bracket. The numerator is equivalent to log4(256)-log4(64) = 4-3 = 1. The Denominator used the addition formula. It is equal to sin(32+13)=sin45=sqt2 over 2. Then get the reciprocal of sqt 2 over 2, which is simply sqt2! Next is the exponent. 2log2(4) = 2x2log2(2)=2x2 = 4. sqt2 raised to 4 is 4. log4(4)=1 and lastly 1-1=0. zero (0) is the final answer :)))