Math help
Someone explain to me this stuff...
1.Express as a single logarithm with a coefficient of 1. Assume that all logarithms in the question have the same base.
4 log z + 5 log y - log w
2. solve for x: 4^x=9
I'm pretty sure i have to convert this one to logarithmic form, correct?
if so then the answer for this would be x=log(subscript4) 9
3. solve for y: log(y^2-1)=log(y+1)
thanks to all that help. :)
LordaeronII
02-12-2004, 19:47
1.Express as a single logarithm with a coefficient of 1. Assume that all logarithms in the question have the same base.
4 log z + 5 log y - log w
2. solve for x: 4^x=9
I'm pretty sure i have to convert this one to logarithmic form, correct?
if so then the answer for this would be x=log(subscript4) 9
3. solve for y: log(y^2-1)=log(y+1)
Well if no base is given actually, then it's the common base, base 10.
Anyways.
1. log(z^4y^5/w)
2. x = log9/log4
3. y^2-1=y+1
y^2-y-2=0
(y-2)(y+1) = 0
y = 2 or y = -1
As much as the answers are appreciated...
I also wanted to know, how to do them?
Can you explain in steps?
LordaeronII
02-12-2004, 20:07
1.Express as a single logarithm with a coefficient of 1. Assume that all logarithms in the question have the same base.
4 log z + 5 log y - log w
Actually I read your question wrong, ignore my answer lol.
okay, 4logz = log(z^4)
5logy = log(y^5)
logw = logw
If you are adding the logs, you can put them into a single log by multiplying, subtracting is division, hence
log[(z^4)(y^5)/w]
2. solve for x: 4^x=9
I'm pretty sure i have to convert this one to logarithmic form, correct?
if so then the answer for this would be x=log(subscript4) 9
Take the log of both sides, you then have
log(4^x) = log9
log(4^x) = xlog4
xlog4 = log 9
Divide both sides by log4
x = log9/log4
3. solve for y: log(y^2-1)=log(y+1)
Okay, well both sides have a common base, log with base 10 (the common base)
Therefore, y^2-1 must = y+1
From there you should be able to solve it.
Therefore, y^2-1 must = y+1
From there you should be able to solve it.
So... would i move it both sides making it
Y^2-y=1-1
y^2-y=0?
LordaeronII
02-12-2004, 20:18
3.
y^2-1=y+1
y^2-y-2=0
(y-2)(y+1) = 0
y = 2 or y = -1
Dear G-d!
No wonder i kept thinking what was wrong!
Sorry... i didnt write the whole question....
log(y^2-1)-1=log(y+1)
Sorry :(
LordaeronII
02-12-2004, 20:31
log(y^2-1)-1=log(y+1)
Hmmmm okee dokee, first express 1 as a common log
1 = log10
So now we have
log(y^2-1)-log10
which equals
log[(y^2-1)/10]
So therefore,
(y^2-1)/10 = y+1
y^2-1 = 10y+10
y^2-10y-11 = 0
(y-11)(y+1) = 0
Therefore, y = 11 or y = -1
thanks i understand'
I think Oo
LordaeronII
02-12-2004, 20:42
Lol no problem, which part aren't you sure about?
I personally just finished doing the logarithms unit so it's fresh in my head.
99.6% on the unit, SO CLOSE!
technically the whole thing you said...
But i;m sure after reviewing your steps, i should be able to understand what you were doing.
right?
Hellshroud
02-12-2004, 20:46
If you are adding the logs, you can put them into a single log by multiplying, subtracting is division
For future questions you may get, remember you can only do that with logs of the same base. Just in case :)
LordaeronII
02-12-2004, 20:49
It might look a bit weird because I have to use ^ rather than actually writing a superscripted number.
It might make more sense to you if you try writing it out by hand with superscripted powers and such.
It might look a bit weird because I have to use ^ rather than actually writing a superscripted number.
It might make more sense to you if you try writing it out by hand with superscripted powers and such.
No.. i don't think it's that.
Probably because i have limited knowledge of this, i am unsure on how it works in general. Thus i dont understand...but looking through everything, although writing it out, might help a bit... it will still take a bit to understand.