Ashish's \(\LaTeX\) playground

$\large \text {This is my note for testing} \LaTeX$ $:P$

$\large \text {A special thanks to}$ $\large \color{#D61F06}{\text {Pi Han Goh}}$ $\large \text {and}$ $\large \color{#3D99F6}{\text {Andrew Ellinor}}$ $\large \text {and}$ $\large \color{#20A900}{\text {Nihar Mahajan}}$$\large \text {for the success of this note.}$ $\begin{aligned} \begin{array}{c}x & = 2+5 \\ & = 7 \end {array} \end{aligned}$

$\begin {pmatrix} 1 & & 0\\ 0 & & 1\ \end {pmatrix}$ × $\begin {pmatrix} 1 & & 0\\ 0 & & 1\ \end {pmatrix}$ = $\begin {pmatrix} 1+0 & & 0+0\\ 0+0 & & 0+1\ \end {pmatrix}$ = $\begin {pmatrix} 1 & & 0\\ 0 & & 1\ \end {pmatrix}$

$\large{\color{magenta}{\text{My name is Ashish}}}$

$\begin{aligned} & x + 7x & = 24\\ \implies & 8x & =24\\ \implies & x & = \dfrac {24}{8}\\ \therefore & x & = \boxed{3} \end{aligned}$

$\ce {{Na}_2CO_3 + H_2O -> NaOH + H_2O + CO_2}$

$\begin{pmatrix} \nearrow \circ \nwarrow & \approx & \nearrow \circ \nwarrow\\ & \bigcup & \\ & \nearrow \searrow \nearrow \searrow \nearrow \searrow & \\& \smile & \end{pmatrix}$

$\LARGE{ \begin{array}{rll} \phantom{0}\ \boxed{{5}} && \\[-2pt] \boxed{{5}}\ \enclose{longdiv}{\boxed{{3}} \ \boxed{7}}\kern-.2ex \\[-2pt] \underline{\boxed{{3}} \ \boxed{{5}}} && \\[-2pt] \boxed{{2}} \end{array} }$

$\LARGE{ \begin{array}{rll} \boxed{{x}}\ + \boxed{{1}} && \\[-2pt] \boxed{{x}}\ + \boxed{{1}} \enclose{longdiv}{\boxed{{x^2}}\ + \boxed{{2x}}\ + \boxed{{1}}}\kern-.2ex \\[-2pt] \underline{\boxed{{x^2}}\ + \boxed{{x}}\ \phantom{0}\ \phantom{0}\ \phantom{0}} \phantom{0}\\[-2pt] \phantom{0}\ \boxed{{x}}\ + \boxed{{1}}\kern-.2ex \\[-2pt] \underline{\boxed{{x}}\ + \boxed{{1}}} && \\[-2pt] \phantom{0}\ \phantom{0}\ \boxed{{0}} \end{array} }$

\begin{aligned} \begin {array} x & \dfrac{29 \times 7}{116}\\ = & \dfrac{{{\cancel {29}}} \times 7}{{\require {cancel}{\cancel {116}}} 4}\\ = & \boxed {\dfrac{7}{4}} \end {array} \end{aligned}

$\huge \text{I ♡ Brilliant}$

$\large \text{Here are some symbols} :- ♤■□●◇◆•○♡♧¡¿《》¤°♧▪$

$\begin{aligned} \dfrac {10}{20} & =\require {cancel}{\dfrac {\xcancel{10}}{\xcancel{20}2}}\\ & = \dfrac {1}{2} \end{aligned}$

${\Huge {\color{#D61F06}{B}}}{\huge{\color{#3D99F6}{R}}}{\LARGE {\color{#CEBB00}{I}}}{\Large{\color{#20A900}{L}}}{\large{\color{magenta}{L}}}{\Large{\color{#69047E}{I}}}{\LARGE{\color{#333333}{A}}}{\huge{\color{#E81990}{N}}}{\Huge{\color{#624F41}{T}}}$

$\begin{aligned} 5+4 & = 8 +1\\ & = 9 \\ \\ 7+2 & = 8+1\\ & = 9 \end{aligned}$

$\Huge \sqrt[\sqrt[x^5]{\sqrt[x^4]{\sqrt[x^3]{\sqrt[x^2]{\sqrt[x]{x}}}}}]{\sqrt[5\sqrt[x]{\sqrt[2x]{\sqrt[3x]{\sqrt[4x]{x}}}}]{\sqrt[5x]{\sqrt[4x]{\sqrt[3x]{\sqrt[2x]{x}}}}}} = x^{x^{5 + \tfrac{x^{11} + 24}{24x^{15}}}}$

Find the real value of $x$ satisfying the real equation above.

$\text{Note}$:- Here $x \neq \{-1 , 0 , 1\}$

$\sqrt[\sqrt[4]{{\left(\sqrt[x]{4x^4}\right)}^{x^4}}]{\sqrt[4]{\sqrt[x]{4}}}$ = $4^{-17x^{-x^3 - 1}}$

$\sqrt[4]{{\left(\sqrt[x]{4x^4}\right)}^{x^4}}$

${\sqrt[4]{\sqrt[x]{4}}}^{\tfrac{1}{\sqrt[4]{{\left(\sqrt[x]{4x^4}\right)}^{x^4}}}}$

$\log_8 \log_4 \color{#3D99F6}{\log_2 16}\\ = \log_8 \log_4 \color{#3D99F6}{4}\\ = \log_8 \color{#D61F06}{\log_4 4}\\ = \log_8 \color{#D61F06}{1}\\ = \color{#20A900}{\log_8 1}\\ = \color{#69047E}{\boxed{0}}$

$\log_{\color{#3D99F6}{b}} \color{#D61F06}{a} = \dfrac{\log \color{#D61F06}{a}}{\log \color{#3D99F6}{b}}$

$\log_{\color{#3D99F6}{\sqrt{2}}} \color{#D61F06}{m}\\ \\ = \dfrac{\log \color{#D61F06}{m}}{\log \color{#3D99F6}{2^{\tfrac{1}{2}}}}\\ \\ = \dfrac{\log m}{\log 2^{\color{#20A900}{\tfrac{1}{2}}}}\\ \\ = \dfrac{\log m}{\color{#20A900}{\dfrac{1}{2}} \log 2}\\ \\ = \dfrac{1}{\frac{1}{2}} \dfrac{\log \color{cyan}{m}}{\log \color{magenta}{2}}\\ \\ = \boxed{2} \log_{\color{magenta}{2}} \color{cyan}{m}$