Question

logx+log (x-1)=log(3x+12) find the value of x​

Answer 1

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