Log questions....really stuck (1 Viewer)

[zenger69]

I'm doing a really weird question in Fitzpatrick...
It goes

"Find a relation between x and y that does not involve logarithms:"

a) log x + log y = log (x+y) answer: y=x/x-1

I keep thinking how the hell did they get that answer?

 

[Captain pi]

LHS = log(x) + log(y)
= log(xy) (A law you should be aware of)
Since RHS = log(x + y)
xy = x + y
xy – y = x
y(x–1) = x
y = x/(x-1) QED
At least, I think that's the answer, you should really use parentheses () if that is the answer.

In a question like this, look for any of the three laws of logarithms that are applicable and see where you can go from them. Have fun!

 

[zenger69]

ok thanks.
i know the laws, it's just a really weird kinda question.

 

[Eagles]

hah, thats crazy! lucky it wasn't in the hsc!

 

[acmilan]

Whenever you get a question involving logs and they ask to find x in terms of y you always need to try and get rid of the logs in some way

 

[zenger69]

hah, thats crazy! lucky it wasn't in the hsc!

I think it was a basic question, just i didn't approach it in the right way......

But yeh now someone has given me the solutions, it seems easy and makes me look like an incompetent 3U student

 

[Captain pi]

hah, thats crazy! lucky it wasn't in the hsc!

I think it was a basic question, just i didn't approach it in the right way......

But yeh now someone has given me the solutions, it seems easy and makes me look like an incompetent 3U student

There, there: *hugs*.

This happens all the time, better to seek advice than to wallow. The familiarity will increase, I assure you.
All of us get stuck some time.

Even you?

No, not me; never.

 

[ando_88]

i have another way of finding that if your interested.

RHS: e^lnx + lny = x + y

e^lnx x e^lny = x + y

xy = x + y (remembering that e^lnx = x)

xy - x - y = 0

y(x-1) = x

y= x/(x-1)

hope that helps!