Question

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

Answer 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) ${\log }_{10}{(1-x)}^2+1-{\log }_{10}{(1+x)}^2=2{\log }_{10}(1-x)$
$\Rightarrow {\log }_{10}(1+x^2)=11+x^2=10\Rightarrow x=\pm 3$