Reply
Fri 8 Jul, 2005 02:02 pm
I have two questions below.
Write each expression as a sum and/or difference of logs. Express powers as factors.
1) LN [x^(2) - x -2/(x+4)^(2) ]^(1/3) =
My work:
LN = log_e (x^2 - x - 2)^1/3 - log_e (x+4) ^1/3(2)
1/3log_e (x^2 - x - 2) - log_e (x+4) ^2/3
Here's my wrong answer:
1/3log_e(x^2-x-2) - 2/3log_e(x+4)
Where did I go wrong?
I don't know what to do with question 2. Here it is:
Write each expression as a sum and/or difference of logs. Express powers as factors.
2) LN 5x(sqrt{ 1-3x})/(x-4)^(3) =
Is it
LN [x^(2) - x -2/(x+4)^(2) ]^(1/3)
or
LN [(x^(2) - x -2)/(x+4)^(2) ]^(1/3)
?
Reply
Is it
LN [x^(2) - x -2/(x+4)^(2) ]^(1/3)
or
LN [(x^(2) - x -2)/(x+4)^(2) ]^(1/3)
?
Question 1 is LN [x^(2) - x -2/(x+4)^(2) ]^(1/3).
Well you solved it as if it were
LN [(x^(2) - x -2)/(x+4)^(2) ]^(1/3)
which is what I suspect it is. Otherwise, I don't think anything can be done.
An additional step you might take is:
since (x^2 - x - 2) = (x-2)(x+1)
ln(x^2 - x - 2) = ln(x-2) + ln(x+1)
LN 5x(sqrt{ 1-3x})/(x-4)^(3)
ln(5x * sqrt(1-3x) / (x-4)^3) =
ln(5) + 0.5*ln(1-3x) - 3*ln(x-4)
Hey
Hey, I like your math steps.
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