More logs!

rtwilton · #1 $Old$ 11-19-2008, 10:06 AM

Can anyone help??

Solve log5(x^2 + x + 4) = 2

and

Solve log4 (x^2 +9) - log4(x+3) = 3

Thanks everyone!

masters · #2 $Old$ 11-19-2008, 10:51 AM

Quote:

Originally Posted by rtwilton $View Post$

Can anyone help??

Solve log5(x^2 + x + 4) = 2

$\log_5(x^2+x+4)=2$

$x^2+x+4=5^2$

Can you finish?

Quote:

Originally Posted by rtwilton $View Post$

Solve log4 (x^2 +9) - log4(x+3) = 3

Thanks everyone!

$\log_4(x^2+9)-\log_4(x+3)=3$

$\log_4\frac{x^2+9}{x+3}=3$

$\frac{x^2+9}{x+3}=4^3$

Can you finish?

rtwilton · #3 $Old$ 11-19-2008, 11:03 AM

I got as far as you did on both, but then I got stuck. I dunno sometimes it's the easy stuff that confuses me...

masters · #4 $Old$ 11-19-2008, 11:10 AM

Quote:

Originally Posted by rtwilton $View Post$

I got as far as you did on both, but then I got stuck. I dunno sometimes it's the easy stuff that confuses me...

$x^2+x+4=25$

$x^2+x-21=0$

This won't factor, so use the quadratic formula to find x.

rtwilton · #5 $Old$ 11-19-2008, 11:11 AM

Ahh the quadratic formula, of course!

rtwilton · #6 $Old$ 11-19-2008, 01:11 PM

So I came up with negative 1/2 plus or minus the square root of 26 divided by 2. Dont know how to write that any better, and can it be simplified any further??

masters · #7 $Old$ 11-19-2008, 01:23 PM

Quote:

Originally Posted by rtwilton $View Post$

So I came up with negative 1/2 plus or minus the square root of 26 divided by 2. Dont know how to write that any better, and can it be simplified any further??

I'm not sure how you managed to come up with that answer.

$x^2+x-21=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-1\pm\sqrt{1^2-4(1)(-21)}}{2(1)}$

$x=\frac{-1\pm\sqrt{85}}{2}$

rtwilton · #8 $Old$ 11-19-2008, 04:38 PM

Well I had almost the same answer as you, see I added the 21 and 4 instead of multiplying!!