logs

**teddyryan** · December 15th 2009, 12:23 AM

how do i solve this equation?

log3(x^2)-log3(3x+2)=0

if i cube the logs does that undo them?

**earboth** · December 15th 2009, 02:28 AM

Originally Posted by teddyryan

how do i solve this equation?

log3(x^2)-log3(3x+2)=0

if i cube the logs does that undo them?

Rewrite the equation:

$\log_3(x^2)-\log_3(3x+2)=0~\implies~\log_3\left(\dfrac{x^2}{3x +2} \right)=0\ ,\ x > -\frac23$

Now use the LHS and the RHS as exponents to the base 3:

$\dfrac{x^2}{3x+2} = 1~\implies~x^2-3x-2=0$

This is a quadratic equation. Use the quadratic formula to solve this equation. You'll get 2 solutions, but be careful: One isn't valid. Why?

**teddyryan** · December 17th 2009, 10:26 AM

it wont be valid if its negative correct? logarithms have a vertical asymptote at 0

**e^(i*pi)** · December 17th 2009, 12:03 PM

Originally Posted by teddyryan

it wont be valid if its negative correct? logarithms have a vertical asymptote at 0

Correct on the negative part but it's a horizontal asymptote at x=0

**HallsofIvy** · December 18th 2009, 05:17 AM

Uhh, no, the y-axis is vertical!

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