# Thread: Expanding logarithm into sums and difference

1. ## Expanding logarithm into sums and difference

Please don't be scared when you see this.

The problem asks me to expand the following logarithm into sums and differences of logarithms whose arguments are as simple as possible:

log[((√x2-1)(x2-x)x3)/(x(x-2)104)]

This is what I did:

= log(√x2-1)+log(x2-x)+3log(x)-log(x)+log(x-2)+log(104)

Not sure if this is correct or if I followed the steps correctly. Would appreciate if someone would look over this and tell me if I did it right.

2. ## Re: Expanding logarithm into sums and difference

Originally Posted by FatimaA

Not sure if this is correct or if I followed the steps correctly. Would appreciate if someone would look over this and tell me if I did it right.
Be very careful with your signs. And we can also go a bit further in the expansion. I'm assuming that that radical is supposed to be $\displaystyle \sqrt{x^2-1}$ and not $\displaystyle \left(\sqrt{x^2}-1\right)$ as you have written:

$\displaystyle \log\left[\frac{\left(\sqrt{x^2-1}\right)\left(x^2-x\right)\left(x^3\right)}{10^4x(x-2)}\right]$.

Start by splitting the fraction:

$\displaystyle = \log\left[\left(\sqrt{x^2-1}\right)\left(x^2-x\right)\left(x^3\right)\right]-\log\left[10^4x(x-2)\right]$

Now split the factors:

$\displaystyle = \left[\log\sqrt{x^2-1}+\log\left(x^2-x\right)+\log x^3\right]-\left[\log10^4+\log x+\log(x-2)\right]$

Notice that the entire denominator ends up getting subtracted. That is, we will have to distribute the -1 to each term on the right side:

$\displaystyle = \log\sqrt{x^2-1}+\log\left(x^2-x\right)+\log x^3-\log10^4-\log x-\log(x-2)$

Notice also that $\displaystyle x^2-x$ can be factored as $\displaystyle x(x-1)$, giving us

$\displaystyle = \log\sqrt{x^2-1}+\log x+\log(x-1)+\log x^3-\log10^4-\log x-\log(x-2)$

And finally, remember that $\displaystyle \sqrt[n]{a^m}=a^{m/n}$. So, we can convert that radical into a rational exponent:

$\displaystyle = \log\left(x^2-1\right)^{1/2}+\log x+\log(x-1)+\log x^3-\log10^4-\log x-\log(x-2)$

Now, can you finish it?

3. ## Re: Expanding logarithm into sums and difference

$\displaystyle \log\left[\frac{\sqrt{x^2-1} \cdot (x^2-x) \cdot x^3}{x \cdot (x-2) \cdot 10^4}\right]$

$\displaystyle \log\left[\frac{\sqrt{x^2-1} \cdot x(x-1) \cdot x^3}{x \cdot (x-2) \cdot 10^4}\right]$

$\displaystyle \log\left[\frac{\sqrt{x-1} \cdot \sqrt{x+1} \cdot (x-1) \cdot x^3}{(x-2) \cdot 10^4}\right]$

$\displaystyle \log\left[\frac{(x-1)^{3/2} \cdot (x+1)^{1/2} \cdot x^3}{(x-2) \cdot 10^4}\right]$

$\displaystyle \frac{3}{2}\log(x-1) + \frac{1}{2}\log(x+1) + 3\log(x) - \log(x-2) - 4$

4. ## Re: Expanding logarithm into sums and difference

Originally Posted by skeeter
$\displaystyle \frac{3}{2}\log(x-1) + \frac{1}{2}\log(x+1) + 3\log(x) - \log(x-2) - 4$
Oh yeah, I forgot to factor the difference of squares there. Thanks skeeter.

5. ## Re: Expanding logarithm into sums and difference

I'm a little confused. You both seemed to do it differently.

Skeeter how did you get log(x-2)-4?

6. ## Re: Expanding logarithm into sums and difference

Originally Posted by FatimaA
I'm a little confused. You both seemed to do it differently.
Not really. They're the same steps, just in a slightly different order. And I didn't finish the last few steps in my post. But it will come out the same if you continue from there.

Originally Posted by FatimaA
Skeeter how did you get log(x-2)-4?
$\displaystyle \log_{10}10^4=4$.

7. ## Re: Expanding logarithm into sums and difference

I thought it was supposed to be simplified as 4log(10)?

8. ## Re: Expanding logarithm into sums and difference

Originally Posted by FatimaA
I thought it was supposed to be simplified as 4log(10)?
Yes. And what is $\displaystyle \log10$?

9. ## Re: Expanding logarithm into sums and difference

Thanks.

Now I don't know how Skeeter got (3/2)log(x-1). I don't understand things if they are not totally spelled out for me lol. I was confused by the whole 3rd step in Skeeter's solution. How did you get (x-1)^3/2?

10. ## Re: Expanding logarithm into sums and difference

Originally Posted by FatimaA
Thanks.

Now I don't know how Skeeter got (3/2)log(x-1). I don't understand things if they are not totally spelled out for me lol. I was confused by the whole 3rd step in Skeeter's solution. How did you get (x-1)^3/2?
Well, $\displaystyle \sqrt{x-1} = (x-1)^{1/2}$. And there's also another factor of $\displaystyle x-1$. So what is $\displaystyle (x-1)^{1/2}\cdot(x-1)$? Just add the exponents.

11. ## Re: Expanding logarithm into sums and difference

Ohhhhh okay I see now. Thank you for clearing that up.

I really love this site.