- Thread starter
- #1

- Thread starter arl2267
- Start date

- Thread starter
- #1

- Admin
- #2

- Thread starter
- #3

- Admin
- #4

- Jan 26, 2012

- 4,093

That's not the equation you should solve.I already solved part of the equation, I'm just not sure what to do once I get to

= x^{2}+3x-1/4=0

When I solve using the quadratic equation after this I end up with:

-3 +/- 2squareroot2/2

$\displaystyle \log_4(x)-\log_4(x+3)=-1$. Using MarkFL's suggestion we simplify this to \(\displaystyle \log_{4}\left( \frac{x}{x+3} \right)=-1\). Using the definition of a logarithm this becomes \(\displaystyle 4^{-1}=\frac{x}{x+3}\).

Can you finish from here?

- Thread starter
- #5

- Admin
- #6

- Jan 26, 2012

- 4,093

Here you can simplify a few ways. Maybe you're familiar with the idea of cross-multiplication of fractions.

\(\displaystyle 1(x+3)=4x\) or simply \(\displaystyle x+3=4x\)

- Thread starter
- #7

- Admin
- #8

- Jan 26, 2012

- 4,093