Solving Logarithmic Equations: log6(x + 5) + log6 x = 2

[jem3]

I need some help solving this logarithmic equation. The resulting quadratic equation doesn't seem to factor easily so I was wondering if it is Ok to solve it by using either the quadrtic formula or by completing the square. Perhaps I've set the problem up wrong and that's why I'm not getting an easily factorable equation. Any suggestions?

This is the problem: log6(x + 5) + log6 x = 2

What I've done so far:

log6 (x + 5) + log6 x = 2
log6 [x(x + 5)] = 2
6^2 = x(x + 5)
36 = x^2 + 5x
0 = x^2 + 5x -36
0 = ( ? ) ( ? )

 

[Deleted member 4993]

Re: Solving Logarithmic Equations

I need some help solving this logarithmic equation. The resulting quadratic equation doesn't seem to factor easily so I was wondering if it is Ok to solve it by using either the quadrtic formula or by completing the square. Perhaps I've set the problem up wrong and that's why I'm not getting an easily factorable equation. Any suggestions?

This is the problem: log6(x + 5) + log6 x = 2

What I've done so far:

log6 (x + 5) + log6 x = 2
log6 [x(x + 5)] = 2
6^2 = x(x + 5)
36 = x^2 + 5x
0 = x^2 + 5x -36
0 = ( ? ) ( ? )

\(\displaystyle x^2 \, + \, 5x \, - \, 36 \, = \, (x \, + \, 9)\cdot (x \, - \,4)\)

Thank you for showing work and posting a succinct question.

To answer your question, it is always acceptable to use "quadratic formula" instead of factoring (I prefer to use quadratic formula - instead of wasting time trying to find factors. In real life problems, rarely you'll meet a nice factorable quadratic equation).

You do realize that although you got two solutions - only one of those is the "valid" solution for your original problem.

 

[jem3]

Thank you for your help, I guess it wasn't that difficult to factor after all. Don't know how I missed it!