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

Let \({\log}_{2\sqrt{3}}\)1728 = x
Then we have \((2\sqrt{3})^x=1728=2^63^3=2^6(\sqrt{3})^6\)
Hence, (2\(\sqrt{3}\))^x = (2\(\sqrt{3}\))⁶
Therefore x = 6. That is, \({\log}_{2\sqrt{3}}1728=6.\)

Multiplying both numerator and denominator by \((\sqrt{6}-\sqrt{2})\), we get
\(\frac{\sqrt{5}}{(\sqrt{6}+\sqrt{2})}=\frac{\sqrt{5}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}=\frac{\sqrt{30}-\sqrt{10})}{4}\)

Given that a and b are the roots of x²- px + q = 0.  Then, a + b = p and ab = q. Thus, \(\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{p}{q}.\) 

We have 3x - 5 ≤ x + 1; which is equivalent to 2x ≤ 6. Hence We have x ≤ 3; The solution set is (-∞, 3].

|2x - 17| = 3, Then, we have 2x - 17 = ±3; which implies x = 10 (or) x = 7

-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.

Observe that \(\sqrt{x^2-10x+25}=\sqrt{(x-5)^2}=|x-5|\)

3|x - 2| + 7 = 19, So that we have, |x - 2| = \(\frac { 19-7 }{ 3 } =4\). Thus, we have either, x - 2 = 4 (or) x - 2 = -4
Therefore the solutions are x = -2 (or) x = 6

y = x² + 6x + 9
Here a = 1, b = 6, c = 9
\(\therefore\) D = b² - 4ac = (6)² - 4 (1) (9)
= 36 - 36 = 0
Since D = 0, the parabola touches the X -axis at only one point.