# Solving equations

• November 13th 2011, 07:25 AM
Fabio010
Solving equations
I dont have any idea how to start this equation.

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

How can i simplify this equation??

(sorry my english)
• November 13th 2011, 07:35 AM
e^(i*pi)
Re: Solving equations
Quote:

Originally Posted by Fabio010
I dont have any idea how to start this equation.

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

How can i simplify this equation??

(sorry my english)

Are you familiar with the laws of logarithms?

$a\ln(k) = \ln(k^a)$

$\ln(ab) = \ln(a) + \ln(b)$

and, because the logarithm is a one-to-one function: $\ln(a) = \ln(b) \Leftrightarrow a = b$
• November 13th 2011, 07:38 AM
Fabio010
Re: Solving equations
never mind.

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

sorry.
• November 13th 2011, 07:41 AM
HallsofIvy
Re: 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.
• November 13th 2011, 07:44 AM
e^(i*pi)
Re: Solving equations
Quote:

Originally Posted by Fabio010
never mind.

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

sorry.

That rule holds for all positive a and b. For example $\ln(e) + \ln(e^3) = \ln(e \cdot e^3) = \ln(e^4) = 4$ which is what we'd expect from adding them separately.
• November 13th 2011, 07:58 AM
Fabio010
Re: Solving equations
So: $ln(x^2-1) + 2ln(2) = ln (4x -1)$

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

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

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

$4x^2-4 = 4x-1$ ?
• November 13th 2011, 08:02 AM
e^(i*pi)
Re: Solving equations
Quote:

Originally Posted by Fabio010
So: $ln(x^2-1) + 2ln(2) = ln (4x -1)$

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

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

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

$4x^2-4 = 4x-1$ ?

Yep, fine thus far. Now solve the quadratic bearing in mind any restrictions on the domain.
• November 13th 2011, 08:14 AM
Fabio010
Re: Solving equations
Ok thanks for the help !
• November 13th 2011, 08:59 AM
e^(i*pi)
Re: Solving equations
Quote:

Originally Posted by Fabio010
Ok thanks for the help !

For reference the answer is $x = \dfrac{3}{2}$