1. ## Solving 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)

2. ## Re: Solving equations

Originally Posted by Fabio010
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)
Are you familiar with the laws of logarithms?

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

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

and, because the logarithm is a one-to-one function: $\displaystyle \ln(a) = \ln(b) \Leftrightarrow a = b$

3. ## Re: Solving equations

never mind.

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

sorry.

4. ## 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.

5. ## Re: Solving equations

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 $\displaystyle \ln(e) + \ln(e^3) = \ln(e \cdot e^3) = \ln(e^4) = 4$ which is what we'd expect from adding them separately.

6. ## Re: Solving equations

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

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

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

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

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

7. ## Re: Solving equations

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

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

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

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

$\displaystyle 4x^2-4 = 4x-1$ ?
Yep, fine thus far. Now solve the quadratic bearing in mind any restrictions on the domain.

8. ## Re: Solving equations

Ok thanks for the help !

9. ## Re: Solving equations

Originally Posted by Fabio010
Ok thanks for the help !
For reference the answer is $\displaystyle x = \dfrac{3}{2}$