logx+log (x-1)=log(3x+12) find the value of x
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Answer:
Given log
x
3+log
3
x=log
3
x+log
3
x
+
2
1
⇒log
x
3+log
3
x=2log
3
x+
2
1
(log
3
x)+
2
1
⇒0=log
3
x−
2
1
(log
3
x)+
2
1
⇒0=
2
2log
3
x−log
3
x+1
⇒0=2log
3
x−log
3
x+1
⇒0=2(
log
3
x
1
)−log
3
x+1
⇒0=2−log
3
2
x+log
3
x
⇒log
3
2
x−log
3
x−2=0
⇒log
3
2
x−2log
3
x+log
3
x−2=0
⇒log
3
x(log
3
x−2)+1(log
3
x−2)=0
⇒(log
3
x+1)(log
3
x−2)=0
⇒log
3
x=−1 and log
3
x=2
⇒x=3
−1
and x=3
2
∴ log
x
3+log
3
x=log
3
x+log
3
x
+
2
1
has 2 solutions
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