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

Question Number 4458 by love math last updated on 29/Jan/16

{l o g}_{3} {(x - 3)}^{2} + {l o g}_{3} ∣ x - 3 ∣= 3

Answered by Yozzii last updated on 29/Jan/16

N o t e t h a t ∣ u ∣= \sqrt{u^{2}} .

∴ {l o g}_{3} ∣ x - 3 ∣= {l o g}_{3} \sqrt{{(x - 3)}^{2}} = 0.5 {l o g}_{3} {(x - 3)}^{2}

∴ T h e g i v e n e q u a t i o n b e c o m e s, w h e r e

n = {l o g}_{3} {(x - 3)}^{2},

n + 0.5 n = 3 \Rightarrow n = 2

\Rightarrow {l o g}_{3} {(x - 3)}^{2} = 2 \Rightarrow {(x - 3)}^{2} = 9

\Rightarrow x = 3 \pm 3 \Rightarrow x = 6 o r x = 0 .

C h e c k i n g, f o r x = 0,

{l o g}_{3} {(0 - 3)}^{2} + {l o g}_{3} ∣ 0 - 3 ∣= {l o g}_{3} 9 + {l o g}_{3} 3 = 3

F o r x = 6,

{l o g}_{3} {(6 - 3)}^{2} + {l o g}_{3} ∣ 6 - 3 ∣= {l o g}_{3} 9 + {l o g}_{3} 3 = 3

Commented by Yozzii last updated on 29/Jan/16

O b s e r v e t h a t i f I h a d d e c i d e d t o

w r i t e n = {l o g}_{3} (x - 3) t h e g i v e n e q u a t i o n

b e c o m e s 2 n + n = 3 \Rightarrow n = 1 \Rightarrow x - 3 = 3 \Rightarrow x = 6 .

{W e}^{'} d b e m i s s i n g t h e s o l u t i o n x = 0

f r o m t h i s m e t h o d .

Commented by Rasheed Soomro last updated on 31/Jan/16

N^{} i c E!