Solving a logarithmic equation.

[thatguythere]

Homework Statement

log₃x+log₃(2x+1)=1
Solve for x.

Homework Equations

The Attempt at a Solution

log₃((2x+1)x)=1
3^log3((2x+1))=3¹
(2x+1)x=3
2x²+x=3
2x²+x-3=0

Using the quadratic formula
x= (-1±√(1²-4(2)(-3)))/(2(2))
= (-1±√(1+24))/(4)
= (-1±√25)/(4)
= (-1±5)/(4)
= 4/4 or -6/4

Is there no solution?

 

[SteamKing]

What's wrong with the two solutions you got from solving the quadratic eqn.?

 

[thatguythere]

Logarithms with negative values are undefined though aren't they? So only 4/4 or 1 would be correct?

 

[Mark44]

Homework Statement

log₃x+log₃(2x+1)=1
Solve for x.

Homework Equations

The Attempt at a Solution

log₃((2x+1)x)=1
3^log3((2x+1))=3¹
(2x+1)x=3
2x²+x=3
2x²+x-3=0

Using the quadratic formula
x= (-1±√(1²-4(2)(-3)))/(2(2))
= (-1±√(1+24))/(4)
= (-1±√25)/(4)
= (-1±5)/(4)
= 4/4 or -6/4

Is there no solution?

You should simplify answers such as these. You have x = 1 or x = -3/2. There is still some work to do, though.

 

[SammyS]

Logarithms with negative values are undefined though aren't they? So only 4/4 or 1 would be correct?

Yes, the other answer is what's called an extraneous solution.

 

[thatguythere]

You should simplify answers such as these. You have x = 1 or x = -3/2. There is still some work to do, though.

If I have x =1 and x =-3/2 and I know that -3/2 is an extraneous answer, then x = 1. I am not certain what else there is to do.

 

[Mark44]

If you've checked your solution and it works, you're done.