Menu Close

log-3-x-3-2-log-3-x-3-3-




Question Number 4458 by love math last updated on 29/Jan/16
log_3 (x−3)^2 +log_3 ∣x−3∣=3
$${log}_{\mathrm{3}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} +{log}_{\mathrm{3}} \mid{x}−\mathrm{3}\mid=\mathrm{3} \\ $$
Answered by Yozzii last updated on 29/Jan/16
Note that ∣u∣=(√u^2 ).  ∴ log_3 ∣x−3∣=log_3 (√((x−3)^2 ))=0.5log_3 (x−3)^2   ∴ The given equation becomes, where  n=log_3 (x−3)^2 ,  n+0.5n=3⇒n=2  ⇒log_3 (x−3)^2 =2⇒(x−3)^2 =9  ⇒x=3±3⇒x=6 or x=0.   Checking, for x=0,  log_3 (0−3)^2 +log_3 ∣0−3∣=log_3 9+log_3 3=3  For x=6,   log_3 (6−3)^2 +log_3 ∣6−3∣=log_3 9+log_3 3=3
$${Note}\:{that}\:\mid{u}\mid=\sqrt{{u}^{\mathrm{2}} }. \\ $$$$\therefore\:{log}_{\mathrm{3}} \mid{x}−\mathrm{3}\mid={log}_{\mathrm{3}} \sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{0}.\mathrm{5}{log}_{\mathrm{3}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\therefore\:{The}\:{given}\:{equation}\:{becomes},\:{where} \\ $$$${n}={log}_{\mathrm{3}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} , \\ $$$${n}+\mathrm{0}.\mathrm{5}{n}=\mathrm{3}\Rightarrow{n}=\mathrm{2} \\ $$$$\Rightarrow{log}_{\mathrm{3}} \left({x}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{2}\Rightarrow\left({x}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\Rightarrow{x}=\mathrm{3}\pm\mathrm{3}\Rightarrow{x}=\mathrm{6}\:{or}\:{x}=\mathrm{0}.\: \\ $$$${Checking},\:{for}\:{x}=\mathrm{0}, \\ $$$${log}_{\mathrm{3}} \left(\mathrm{0}−\mathrm{3}\right)^{\mathrm{2}} +{log}_{\mathrm{3}} \mid\mathrm{0}−\mathrm{3}\mid={log}_{\mathrm{3}} \mathrm{9}+{log}_{\mathrm{3}} \mathrm{3}=\mathrm{3} \\ $$$${For}\:{x}=\mathrm{6},\: \\ $$$${log}_{\mathrm{3}} \left(\mathrm{6}−\mathrm{3}\right)^{\mathrm{2}} +{log}_{\mathrm{3}} \mid\mathrm{6}−\mathrm{3}\mid={log}_{\mathrm{3}} \mathrm{9}+{log}_{\mathrm{3}} \mathrm{3}=\mathrm{3} \\ $$
Commented by Yozzii last updated on 29/Jan/16
Observe that if I had decided to   write n=log_3 (x−3) the given equation  becomes 2n+n=3⇒n=1⇒x−3=3⇒x=6.  We′d be missing the solution x=0  from this method.
$${Observe}\:{that}\:{if}\:{I}\:{had}\:{decided}\:{to}\: \\ $$$${write}\:{n}={log}_{\mathrm{3}} \left({x}−\mathrm{3}\right)\:{the}\:{given}\:{equation} \\ $$$${becomes}\:\mathrm{2}{n}+{n}=\mathrm{3}\Rightarrow{n}=\mathrm{1}\Rightarrow{x}−\mathrm{3}=\mathrm{3}\Rightarrow{x}=\mathrm{6}. \\ $$$${We}'{d}\:{be}\:{missing}\:{the}\:{solution}\:{x}=\mathrm{0} \\ $$$${from}\:{this}\:{method}. \\ $$
Commented by Rasheed Soomro last updated on 31/Jan/16
N^  icE!
$$\mathcal{N}^{\:} {ic}\mathcal{E}! \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *