Question

Match the possible values of x mentioned in column II for the equations mentioned in column I

	Column I		Column II
(A)	$x^{(lo{g}_{5}x{)}^2+5lo{g}_{5}x+2}=5^8$	(p)	1
(B)	$(x^2+6{)}^{lo{g}_{3}x}=(5x{)}^{lo{g}_{3}x}$	(q)	2
(C)	$(3+2\sqrt{2}{)}^{x^2-6x+9}+(3-2\sqrt{2}{)}^{x^2-6x+9}=6$	(r)	3
(D)	$lo{g}_8\frac{8}{x^2}÷(lo{g}_{8}x{)}^2=3$	(s)	4
		(t)	5

Answer 1

(A)
$x(lo{g}_{5}x{)}^2+5lo{g}_{5}x+2=5^8$
$\Rightarrow lo{g}_x{5}^8=(lo{g}_{5}x{)}^2+5lo{g}_{5}x+2$
$\Rightarrow 8lo{g}_{x}5=(lo{g}_{5}x{)}^2+5lo{g}_{5}x+2(\text{Let}lo{g}_{5}x=A)$
$\Rightarrow \frac{8}{A}=A^2+5A+2$
$\Rightarrow A^2+5A^2+2A-8=0$
$\Rightarrow (A-1)(A+2)(A+4)=0$
$\Rightarrow x=5,5^{-2},5^{-4}$
Solution $A\left(t\right)$

(B)
$(x^2+6{)}^{{\log }_{3}x}=(5x{)}^{lo{g}_{3}x}$
$\Rightarrow x^2+6=5x$
$\Rightarrow x^2–5x+6=0$
$\Rightarrow x^2–3x–2x+6=0$
$\Rightarrow (x–2)(x–3)=0$
$\therefore \begin{array}{c}x-2=0\\ x=2\end{array}|\begin{array}{c}x-3=0\\ x=3\end{array}$
When x = 1, the equation is satisfied. So, solution is p, q, r.

(C)
$(3+2\sqrt{2}{)}^{x^2-6x+9}+(3-2\sqrt{2}{)}^{x^2-6x+9}=6$
Let $(3+2\sqrt{2}{)}^{x^2-6x+9}=A$
$(3-2\sqrt{2}{)}^{x^2-6x+9}=\frac{(3-2\sqrt{2}{)}^{x^2-6x+9}}{(3+2\sqrt{2}{)}^{x^2-6x+9}}\times (3+2\sqrt{2}{)}^{x^2-6x+9}$
$=\frac{1}{(3+2\sqrt{2}{)}^{x^2-6x+9}}=\frac{1}{A}$
Now,
$\Rightarrow A+\frac{1}{A}=6$
$\Rightarrow A^2+1=6A$
$\Rightarrow A^2-6A+1=0$
$\therefore A=\frac{6\pm \sqrt{32}}{2}$
$A=\frac{6\pm 4\sqrt{2}}{2}=\frac{2(3\pm 2\sqrt{2})}{2}$
$=(3+2\sqrt{2}),(3-2\sqrt{2})$
$\therefore A=3+2\sqrt{2}A=3-2\sqrt{2}$
$\Rightarrow (3+2\sqrt{2}{)}^{x^2-6x+9}=(3+2\sqrt{2}{)}^{-1}$
$(3+2\sqrt{2}{)}^{x^2-6x+9}=(3+2\sqrt{2}{)}^{-1}$
$\therefore x^2-6x+9=1x^2-6x+9=-1$
$\Rightarrow x^2-6x+8=0x^2-6x+10=0$
$\Rightarrow x^2-4x-2x+8=0$
$\Rightarrow x(x-4)-2(x-4)=0=\frac{6\pm \sqrt{-4}}{2}$
$\Rightarrow (x-4)(x-2)=0$
$\Rightarrow x=2,4$
Solution of C (q, s)

(D)
$lo{g}_8\frac{8}{x^2}(lo{g}_{8}x{)}^2=9$
$\Rightarrow \frac{1-lo{g}_8{X}^2}{(lo{g}_{8}X{)}^2}=3$
$\Rightarrow \frac{1-2y}{y^2}=3(\text{Let}lo{g}_{8}x=y)$
$\Rightarrow 3y^2+2y-1=0$
$\Rightarrow 3y^2+2y-1=0$
$\Rightarrow 3y(y+1)-(y+1)=0$
$\therefore y=\frac{1}{3},-1$
$y=\frac{1}{3}$
$lo{g}_{8}x=\frac{1}{3}$
$\therefore 8^{\frac{1}{3}}=x$
$y=-1$
$\Rightarrow x=\frac{1}{8}$
$\therefore 8^{\frac{1}{3}}=x$
$x=2$
Solution of D (q)