Hey guys so i'm trying to turn this

juctommaccedo662f

juctommaccedo662f

Answered question

2022-03-24

Hey guys so i'm trying to turn this equation into it's sum/difference logarithm. However, the part that messes me up is turning the bottom of the fraction
log(x2+2x+1x23x+2)2.
I think it will turn into this:
4log(x+1)2(log(x1)+log(x2)),
but I don't want it to be (x1)2 times (x2)2 . How do I solve it so it's ((x1)(x2))2?

Answer & Explanation

Alejandra Hanna

Alejandra Hanna

Beginner2022-03-25Added 10 answers

I assume you want to simplify
log((x2+2x+1x23x+2)2)
First recall the following properties of logarithm.
1. log(am)=mlog(a)
2. log(ab)=logalogb
3. log(ab)=loga+logb
Using the first property, we get that
log((x2+2x+1x23x+2)2)=2log(x2+2x+1x23x+2)
Now using the second property, we get that
log(x2+2x+1x23x+2)=log(x2+2x+1)log(x23x+2)
Next note that (x2+2x+1)=(x+1)2 and x23x+2=(x1)(x2). Hence, we have that
log(x2+2x+1)=log((x+1)2)=2log(x+1)
using property (1). Similarly, we have that
log(x23x+2)=log((x2)(x1))=log(x2)+log(x1)
using property (3). Putting all these together, we get what you want.

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