I dont have any idea how to start this equation.

$\displaystyle ln(x^2-1) + 2ln(2) = ln (4x -1)$

How can i simplify this equation??

(sorry my english)

Printable View

- Nov 13th 2011, 07:25 AMFabio010Solving equations
I dont have any idea how to start this equation.

$\displaystyle ln(x^2-1) + 2ln(2) = ln (4x -1)$

How can i simplify this equation??

(sorry my english) - Nov 13th 2011, 07:35 AMe^(i*pi)Re: Solving equations
- Nov 13th 2011, 07:38 AMFabio010Re: Solving equations
never mind.

I dont know why but in my head ln(a) + ln (b) = ln (ab) if a = b...

sorry. - Nov 13th 2011, 07:41 AMHallsofIvyRe: Solving equations
What you do need to think about is that logarithm is a "one to one" function. That is, ln(a)= ln(b) if and only if a= b.

- Nov 13th 2011, 07:44 AMe^(i*pi)Re: Solving equations
- Nov 13th 2011, 07:58 AMFabio010Re: Solving equations
So: $\displaystyle ln(x^2-1) + 2ln(2) = ln (4x -1)$

$\displaystyle ln(x^2-1) + ln(4) = ln (4x -1)$

$\displaystyle ln ((x^2-1)4)) = ln(4x -1)$

$\displaystyle ln ((4x^2-4)) = ln (4x-1)$

$\displaystyle ln ((4x^2-4)) - ln (4x-1) = 0 $

$\displaystyle ln ((\frac{4x^2-4}{4x-1})) = 0$

$\displaystyle \frac{4x^2-4}{4x-1} = e^0$

$\displaystyle 4x^2-4 = 4x-1$ ? - Nov 13th 2011, 08:02 AMe^(i*pi)Re: Solving equations
- Nov 13th 2011, 08:14 AMFabio010Re: Solving equations
Ok thanks for the help !

- Nov 13th 2011, 08:59 AMe^(i*pi)Re: Solving equations