We are given log_a 324 = 4, which gives
a⁴ = 324 = 3⁴(\(\sqrt{2}\))⁴. Therefore a = 3\(\sqrt{2}\)

-x² + 3x + 1 = 0
Given equation is -x² + 3x + 1 = 0
Here a = -1, b = 3, c = 1
\(\therefore\) D = b² - 4ac = 3² - 4 (-1) (1)
= 9 + 4 = 13
Since D > 0, the two roots are real and distinct.

Since A\(\times\) A contains 16 elements, then A must have 4 elements
\(\Rightarrow\) n(A) = 4.
The elements of A \(\times\) A are (1, 3) and (0, 2)
\(\therefore\) The possibilities of elements of A are {0, 1, 2, 3}

[a < b ⇒ ay > by for all y< 0]
Here y = -1
Given |3-x| < 7
This means -7 < 3 -x < 7
⇒-7-3 < -x < 7-3
⇒-10 < -x < 4
⇒10 > x > -4
\(\therefore\) The Solution set is \(x\in \left( -\infty ,-4 \right) \cup \left( -4,10 \right) \)

{x \(\in \)​ ​​​​​​N : 4x + 9 < 52}
Let C = {x\(\in \)N:4x + 9 < 52}
\(\Rightarrow\) C = {x\(\in \)N:4x < 52 - 9}
\(\Rightarrow\) C = {x\(\in \)N: 4x < 43}
\(\Rightarrow\) C = \(\left\{ x\in N:x<\frac { 43 }{ 4 } \right\} \) \(\Rightarrow\) C = {x \(\in \) N : x < 10.75}
\(\Rightarrow\) C = {1,2,3,4,5,6,7,8,9,10}.

= \(\left( { -10 }^{ 3\times \frac { -2 }{ 3 } } \right) =-{ 10 }^{ -2 }=\frac { -1 }{ { 10 }^{ 2 } } =\frac { -1 }{ 100 } \)

We know \(|x|=\begin{cases} -x\ \ if\ x\le0 \\ x\quad if\ x>0 \end{cases}\)
So, \(f(x)=\begin{cases} 2x-(-x)\quad if x \le 0\\ 2x-x\quad if\ x>0 \end{cases}\)
Thus, \(f(x)=\begin{cases}3x\quad ifx\le0\\x\quad if\ x>0 \end{cases}\)
Also, \(g(x)=\begin{cases} 2x+(-x)\quad if \ x\le 0\\2x+x\quad if\ x>0 \end{cases}\)
Thus, \(g(x)=\begin{cases} x\quad if x \le 0\\3x\quad if x >0 \end{cases}\)
Let \(x\le0.\) Then 
(f o g) (x) = f(g{x)) = f(x) = 3x.,
The last equality is taken because \(3x\le0\) whenever \(x\le0.\)
Let x > 0. Then
(f o g)(x) = f(g(x)) = f(3x) = 3x.
Thus (f o g)(x) = 3x for all x.

Since the degree of the numerator is equal to the degree of the denominator, let us divide the numerator by the denominator

\(∴\ \ {x^2+x+1\over x^2-5x+6}=1+{6x-5\over x^2-5x+6 }\)
Consider \({6x-5\over x^2-5x+6}={6x-5\over (x-3)(x-2)}={A\over (x-3)}+{B\over (x-2)}\)
\(⇒\ {6x-5\over x^2-5x+6}={A(x-2)+B(x-3)\over (x-3)(x-2}\)
⇒ 6x - 5 = A(x - 2) + B(x - 3)
Putting x=2 in (2) we get
7 = B(-1) ⇒ B = -7
Putting x = 3 in (2) we get
13 = A(1) ⇒ A = 13
\(∴\ {6x-5\over x^2-5x+6}={13\over x-3}-{7\over x-2}\)
Substituting in (1) we get
\({x^2+x+1\over x^2-5x+6}=1+{13\over x-3}-{7\over x-2}\)

LHS = \(log2+16log{16\over 15}+12log{25\over 24}+7log{81\over 80}\)
\(=log2+log\left(16\over 15\right)^{16}+log\left(25\over 24\right)^{12}+log \left(81\over 80\right)^7\)
\(=log2\times{(2^4)^{16}\over (3\times5)^{16}}\times{(5^2)^{12}\over (2^2\times3)^{12}}\times{(3^4)^7\over 2^{28}\times5^7}\)
\(=log2^1\times{2^{64}\over 3^{16}}\times{5^{24}\over 2^{36}\times3^{12}}\times{3^{28}\over 2^{28}\times5^7}\)
\(=log{2^{1+64}.5^{24}.3^{28}\over 3^{16+12}.5^{16+7}.2^{36+28}}\)  \(\left[∵\ {a^m\over a^n}=a^{m-n} \right]\)

= log 2^65-64 x 5^24-23 = log 2¹ \(\times\) 5¹ = log₁₀¹⁰ = 1 = RHS

Given \(\frac { 1 }{ 3-\sqrt { 8 } } -\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } +\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } -\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } -2 } \)   ..(1)
Multiplying each term by the conjugate of the denominator we get
 \(\frac { 1 }{ 3-\sqrt { 8 } } \) = \(\frac { 1 }{ 3-\sqrt { 8 } } \times \frac { 3+\sqrt { 8 } }{ 3+\sqrt { 8 } } =\frac { 3+\sqrt { 8 } }{ { 3 }^{ 2 }-\sqrt { 8 } ^{ 2 } } =\frac { 3+\sqrt { 8 } }{ 9-8 } =3+\sqrt { 8 } \)
\(\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } \)= \(\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } \times \frac { \sqrt { 8 } +\sqrt { 7 } }{ \sqrt { 8 } +\sqrt { 7 } } =\frac { \sqrt { 8 } +\sqrt { 7 } }{ 8-7 } =\frac { \sqrt { 8 } +\sqrt { 7 } }{ 1 } =\sqrt { 8 } +\sqrt { 7 } \)
\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \) =\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \times \frac { \sqrt { 7 } +\sqrt { 6 } }{ \sqrt { 7 } +\sqrt { 6 } } =\frac { \sqrt { 7 } +\sqrt { 6 } }{ \sqrt { 7 } ^{ 2 }+\sqrt { 6 } ^{ 2 } } =\frac { \sqrt { 7 } +\sqrt { 6 } }{ 7-6 } =\sqrt { 7 } +\sqrt { 6 } \)
\(\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } \)  = \(\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } \times \frac { \sqrt { 6 } +\sqrt { 5 } }{ \sqrt { 6 } +\sqrt { 5 } } =\frac { \sqrt { 6 } +\sqrt { 5 } }{ \sqrt { 6 } ^{ 2 }+\sqrt { 5 } ^{ 2 } } =\frac { \sqrt { 6 } +\sqrt { 5 } }{ 6-5 } \)
 \(\frac { 1 }{ \sqrt { 5 } -2 } \) =  \(\frac { 1 }{ \sqrt { 5 } -2 } \times \frac { \sqrt { 5 } +2 }{ \sqrt { 5 } +2 } =\frac { \sqrt { 5 } +2 }{ \sqrt { 5 } ^{ 2 }+2^{ 2 } } =\frac { \sqrt { 5 } +2 }{ 5-4 } =\sqrt { 5 } +2\)
Substituting all these values in (1)we get
\(\frac { 1 }{ 3-\sqrt { 8 } } -\frac { 1 }{ \sqrt { 8 } -\sqrt { 7 } } +\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } -\frac { 1 }{ \sqrt { 6 } -\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } -2 } \) = 5

Given f(x) = 3x - 4
Let y = 3x - 4 ⇒ y + 4 = 3x
\(⇒ x={y+4\over 3}\)
Let g(y) = \(y+4\over 3\)
Now gof(n) = g(f(n)) = g(3\(\times\) -4) = \({3x-4+4\over 3}={3x\over 3}=x\)
and fog(y) = f(g(y)) = \(f\left(y+4\over 4\right)=3\left(y+4\over 3\right)-4=y+4-4=y\)
Thus, gof(x) = I_x and fog (y) = I_y
This implies that f and g are bijections and inverses to each other
Hence f is bijection and \(f^{-1} (x)={y+4\over 3}\)
Replacing y by x, we get f^-1 (x) = \(\frac { x+4 }{ 3 } \)

Hence, the graph of y = f^-1(x) is the reflection of the graph of f in y = x

Note that x > 0.
log₂x + log₄x + log₁₆x = \(\frac{7}{2}\) becomes \(\frac{1}{log_x2}+\frac{1}{log_x4}+\frac{1}{log_x16}=\frac{7}{2}\) (change of base rule)
Thus \(\frac{1}{a}+\frac{1}{2a}+\frac{1}{4a}=\frac{7}{2}\) where a = log_x 2. That is \(\frac{7}{4a}=\frac{7}{2}\)
Thus, a = \(\frac{1}{2}\) and so, log_x 2 =\(\frac{1}{2}\) which gives \(x^{\frac{1}{2}}\)= 2
Thus, x = 2² = 4

(g o f) (x) = g(f(x)) = g(3x - 4) = (3x-4)²+3 = 9x²-24x+19
(f o g)(x) = f(g(x)) = f(x²+ 3) = 3(x²+ 3)-4 = 3x²+5.
Thus the operation "Composition of functions" is in general. (not commutative)

If x < -3 or x > 3, then x² will be greater than 9 and hence 9 - x² will become negative which has no square root in R.
So x must lie on the interval [- 3, 3].
Also if \(x\ge-1\) or \(x\le 1,\) then x²-1 will become negative or zero. If it is negative, x² - 1 has no square root in R.  If it is zero, f is not defined. So, x must lie outside [- 1, 1]. 
That is x must lie on \(( -\infty,-1 ]\cup[1,\infty),\) Combining these two conditions, the largest possible domain for f is \([-3,3]\cap((-\infty, -1)\cup(1,\infty)).\)  That is \([-3,-1)\cup(1, 3].\)

The given equation is equivalent to the system (x + 4) ≥ 0 and 6 - 4x - x² = (x + 4)²
This implies x ≥ -4 and x² + 6x + 5 = 0. Thus x = -1, -5.
But only x = -1 satisfies both the conditions. Hence, x = -1.

Let f(x) = \(\frac { 1 }{ 1-2sinx } \)
When the denominator is 0,
1-2 sin x = 0
\(\Rightarrow\) 1 = 2 sin x
\(\Rightarrow sin\quad x=\frac { 1 }{ 2 } \)
\(\Rightarrow sin\quad x=sin\frac { \pi }{ 6 } \)
\(\Rightarrow x=n\pi +{ (-1) }^{ n }\frac { \pi }{ 6 } n\in Z\)   \(\left[ \because sin\quad x=sin\alpha \Rightarrow x=n\pi +{ (-1) }^{ n }\alpha \quad n\in Z \right] \)
Domain of f(x) is R - \(\left( n\pi +{ (-1) }^{ n }\frac { \pi }{ 6 } \right) ,n\in Z\)

Given inequality is 2x² + x - 15 < 0.
On factorising we get,
(x + 3)(2x - 5) < 0
\(⇒\ 2(x+3)\left(x-{5\over 2}\right)\le0\)

The critical points are \(-3,{5\over 2}.\) where the factors Vanish.
Draw the number line and mark the critical points.
The possible intervals are \((-\infty ,-3)\left(-3,{5\over 2}\right)\left({5\over 2},\infty\right)\)

The in equality 2x² +x - 15 < 0 is satisfied only in the interval \(\left[-3,{+5\over 2}\right]\)
∴ Solution set is \(\left[-3,{+5\over 2}\right]\)

Given a²+ b² = 7ab
Adding 2ab both sides we get,
a²+b²+2ab = 7ab + 2ab
⇒ (a+b)² = 9ab
⇒ \(\frac { (a+b)^{ 2 } }{ 9 } \)  = ab
⇒  \(\left( \frac { a+b^{ 2 } }{ 9 } \right) ^{ 2 }\)  = ab
Taking square root,we get
\(\frac { a+b }{ 3 } \)  = ab
\(log\left( \frac { a+b }{ 3 } \right) =log(ab)^{ \frac { 1 }{ 2 } }\)
= \(\frac { 1 }{ 2 } \) log (ab)
log \(\left( \frac { a+b }{ 3 } \right) \) = \(\frac { 1 }{ 2 } \) [log a + log b]
Hence proved.