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Logarithms

I really struggle on logarithms :frown:

2logaN-loga(3N-4)=loga2

help please :smile:)
Reply 1
Avatar for davros
davros
16
Original post by strawbericalpony
I really struggle on logarithms :frown:

2logaN-loga(3N-4)=loga2

help please :smile:)


Can you use your log rules to help you rewrite the 1st term on the left, and then combine it with the second term?
Original post by davros
Can you use your log rules to help you rewrite the 1st term on the left, and then combine it with the second term?


Yeah I've got as far as

logaN^2 - loga(3N-4) = loga2

logaN^2/3n-4 = loga2

and then I'm confused
Reply 3
Avatar for davros
davros
16
Original post by strawbericalpony
Yeah I've got as far as

logaN^2 - loga(3N-4) = loga2

logaN^2/3n-4 = loga2

and then I'm confused


OK, that's great so now you have:

log (something) = log(something else)

so you must have

something = something else

i.e. you can remove the logarithms (technically you're raising a to the power of both sides), and that gives you an equation involving N which you can rearrange to get a quadratic :smile:
Reply 4
Avatar for Zakee
Zakee
After removing logarithms, (cancelling by loga loga ) on both sides.

[br][br]n2/(3n−4)=2[br][br][br][br]n^2/(3n-4) = 2[br][br]

It becomes an effortless quadratic which can be solved any way you please. :smile:
(edited 10 years ago)

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