not entirely sure what to do... anyone thing they could help?
log4(5-x) - log4(3-x) = 2 x<3
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Paper 2 2005
Started by laura_lou92, May 20 2009 02:52 PM
3 replies to this topic
#2
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Posted 20 May 2009 - 03:15 PM
QUOTE (laura_lou92 @ May 20 2009, 03:52 PM) <{POST_SNAPBACK}>
not entirely sure what to do... anyone thing they could help?
log4(5-x) - log4(3-x) = 2 x<3
log4(5-x) - log4(3-x) = 2 x<3
Haha i just did this one 5 minutes ago
Right, you already know log_{4}(5-x)-log_{4}(3-x)=2
So you re-arrange to log_{4}(5-x)-log_{4}(3-x)=log_{4}4^2 Remember : log_{4}4=1
Then you can cancel out the logs:
(5-x)/(3-x)=4^2
16(3-x)=(5-x)
48-16x=5-x
5-15x=48
-15x=43
x=-43/15
Hope this helped! Any questions just reply back.
Yeah!
#3
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Posted 20 May 2009 - 03:40 PM
QUOTE (Philip @ May 20 2009, 04:15 PM) <{POST_SNAPBACK}>
QUOTE (laura_lou92 @ May 20 2009, 03:52 PM) <{POST_SNAPBACK}>
not entirely sure what to do... anyone thing they could help?
log4(5-x) - log4(3-x) = 2 x<3
log4(5-x) - log4(3-x) = 2 x<3
Haha i just did this one 5 minutes ago
Right, you already know log_{4}(5-x)-log_{4}(3-x)=2
So you re-arrange to log_{4}(5-x)-log_{4}(3-x)=log_{4}4^2 Remember : log_{4}4=1
Then you can cancel out the logs:
(5-x)/(3-x)=4^2
16(3-x)=(5-x)
48-16x=5-x
5-15x=48
-15x=43
x=-43/15
Hope this helped! Any questions just reply back.
#4
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Posted 20 May 2009 - 03:44 PM
QUOTE (laura_lou92 @ May 20 2009, 04:40 PM) <{POST_SNAPBACK}>
QUOTE (Philip @ May 20 2009, 04:15 PM) <{POST_SNAPBACK}>
QUOTE (laura_lou92 @ May 20 2009, 03:52 PM) <{POST_SNAPBACK}>
not entirely sure what to do... anyone thing they could help?
log4(5-x) - log4(3-x) = 2 x<3
log4(5-x) - log4(3-x) = 2 x<3
Haha i just did this one 5 minutes ago
Right, you already know log_{4}(5-x)-log_{4}(3-x)=2
So you re-arrange to log_{4}(5-x)-log_{4}(3-x)=log_{4}4^2 Remember : log_{4}4=1
Then you can cancel out the logs:
(5-x)/(3-x)=4^2
16(3-x)=(5-x)
48-16x=5-x
5-15x=48
-15x=43
x=-43/15
Hope this helped! Any questions just reply back.
Thank u so much =)
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