Question

Match the equations in List 1 with the solutions in List 2

	List I		List II
A.	${\log }_{0.25}{(x^2+2x-8)}^2$ $-{\log }_{0.5}(10+3x-x^2)=1$	1.	$\{-3\}$
B.	${\log }_2(x^2+7)$ $=5+{\log }_{2}x$ $-\frac{6}{{\log }_2(x+\frac{7}{x})}$	2.	$\left\{\frac{1}{4}\right\}$
C.	${\log }_{(1-2x)}(6x^2-5x+1)$$-{\log }_{(1-3x)}(4x^2-4x+1)=2$	3.	$\{\frac{1}{6}(\sqrt{313}-1),\frac{1}{2}(\sqrt{73}-7)\}$
D.	${\log }_{10}(1{+x}^2-2x)+1-{\log }_{10}(1+x^2)=$ $2{\log }_{10}(1-x)$	4.	$\{1,7\}$

• A. A- 3, B- 4, C- 2, D-1
• B. A- 3, B- 4, C- 1, D-2
• C. A- 3, B- 1, C- 2, D-4
• D. A- 3, B- 2, C- 1, D-4

Answer 1

Answer: (A)

A- 3, B- 4, C- 2, D-1

(a) Rewrite as
${\log }_{0.5}\frac{|x^2+2x-8|}{|10+3x-x^2|}=1\Rightarrow \frac{|x^2+2x-8|}{10+3x-x^2}=\frac{1}{2}$
$\Rightarrow 2|x^2+2x-8|=|10+3x-x^2|$
$x\in \{\frac{1}{6}(\sqrt{313}-1),\frac{1}{2}(\sqrt{73}-7)\}$
(b) Put $t={\log }_2(x^2+7)-{\log }_{2}x$ to obtain
$t=5-\frac{6}{t}\Rightarrow t^2-5t+6=0\Rightarrow t=2,3$
$\Rightarrow {\log }_2\left(\frac{x^2+7}{x}\right)=2,3\Rightarrow \frac{x^2+7}{x}=4,8\Rightarrow x=1,7$
(c) Given equation is valid when $1-2x>0,1-2x\ne 1,1-3x>0,1-3x\ne 1,6x^2-5x+1>0,4x^2-4x+1>0$
Rewrite the equation as
${\log }_{(1-2x)}\left[\right(1-2x\left)\right(1-3x\left)\right]-{\log }_{1-3x}{(1-2x)}^2=2\Rightarrow 1+t-\frac{2}{t}=2$
$t=\frac{\log (1-3x)}{\log (1-2x)}\Rightarrow t^2-t-2=0\Rightarrow t=-1,2\Rightarrow x=\frac{1}{4}$
(d) ${\log }_{10}{(1-x)}^2+1-{\log }_{10}{(1+x)}^2=2{\log }_{10}(1-x)$
$\Rightarrow {\log }_{10}(1+x^2)=11+x^2=10\Rightarrow x=\pm 3$