Solving for x: "1/2*log(5)(x^2-1)=1/4+1/2log(5)(x-1)

[parabol]

Homework Statement

1/2*log(5)(x^2-1)=1/4+1/2log(5)(x-1)

I have put the base to the log in the first set of brackets following log.

The question is simply solve for x the equation

The Attempt at a Solution

I firstly moved the logs onto the smae side and then multiplied both sides by1/2 (not shown) thus giving

log(5)(x^2-1)-log(5)(x-1)=1/2

log(5)((x^2-1)/(x-1))=1/2

Am i right in the saying the follwoing is the next step?

(x2-1)/(x-1)=5^(1/2)

If I am then can someone please help me understand how I am supposed to simplify the equaiton down to give a single x.

Thanks in advance.

 

[Bohrok]

Assuming that log(5) is log₅, that is right. How can you rewrite
$$\frac{x^2-1} {x-1}$$

specifically the numerator?

 

[parabol]

That was exactly what I meant with regards to the base of the log.

I'm sorry I don't follow. I think I am looking to hard into this problem as I can see no way of simplifying (x²-1)/(x-1)

 

[Bohrok]

Do you know how to factor x² - 1?

 

[HallsofIvy]

You have
$$\frac{x^2- 1}{x- 1}= \sqrt{5}$$

What Bohrok is suggesting is that you factor the numerator. There is a simple cancelation.

 

[parabol]

All sorted now.

Thanks very much for the sanity checks with this.