LOG fractions: Let Log 2=a, log 3=b & log 5=c, express the log terms in a,b & c

H

helpmepls!

New member
Joined
Feb 18, 2017
Messages
13
LOG fractions: Let Log 2=a, log 3=b & log 5=c, express the log terms in a,b & c

the problem is Log 6/25 (its a fraction, idk how to do those on here). I got the answer but then I was told it is wrong because the question actually is..

"Let Log 2=a, log 3=b & log 5=c, express the log terms in a,b & c" how do I go about that? Thank You!
 
What are your thoughts? What have you tried? Please comply with the rules as laid out in the Read Before Posting thread that's stickied at the top of each sub-forum, and share with us any and all work you've done on this problem, even the parts you know for sure are wrong. Thank you.

As a hint, you may want to review the rules of logarithmshttp://www.rapidtables.com/math/algebra/Logarithm.htm, in particular noting that log(a/b) = log(a) - log(b) and log(ab) = log(a) + log(b).
 
so far all I got is Log 6/25 is equal to log(6)-log(25). Log(25) converts to 2log(5). Log 6 isn't found with a power so I just left it as it. So now I have Log(6)-2Log(5), which would convert to Log(6)-2(c). I don't know if that's the correct answer, but so far I just have the C
 
helpmepls! said:
It doesn't have to be in power form??
Click to expand...
If by "It" you're talking about 6, then no. The 6 does not need to be written as a power.

It would be handy to write 6 as a product. Then you would have:

log(2*3) - 2*log(5)

Earlier, you were told that this property would be useful:

log(a*b) = log(a) + log(b)

Try applying this property to log(2*3), and see what happens. :cool:
 
You must log in or register to reply here.
Top