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Article #41

hard problems

#41
From: Mr. X
Date: Mon, 23 Feb 1998 00:00
31 lines
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Here are some hare math problems that I've been working on.  I've had
tremendous difficulty with them.  If you figure some of them out, please
e-mail me with the answer and the work used to find the answer at
boot-leg@ix.netcom.com .  Here they are:
1)  A certain virus has infected 10,000 people in a geographical area.  The
treatment for the virus can reduce the number of infected peopleby 10% each
week, so the decrease is exponential.  The following data has been collected
for the first three weeks.  How many weeks will pass until only 10% of the
original 10,000 have the virus?
    (week #, # of people infected): (0,10000); (1,9000); (2,8100); (3,7300)
2)  Write each as a single logarithm.
        a)  3ln x + ln(pie) - 2ln(x-2)
        b)  (1/2)ln x + (1/2)ln y - ln(x-5)
3)  If we define (2n +1)! to mean the product of (1)(2)(3). . .(2n + 1) and
(2n + 1)!! to mean the product of (1)(3)(5). . .(2n + 1), express (2n + 1)!!
in terms of (2n + 1)!
4)  If (ax)^log a  = (bx)^log b , a,b>0, a is not equal to b, a is not equal
to 1, b is not equal to 1, and the logarithmic base is the same throughout,
express x in terms of a and b
5)  If f is a function of x only, and q is a function of y only, determine f
and g suck that the log f + log g = log(1+z) where z = x + xy + y

In this last problem there are logs with different bases than ten, so I will
be using this format:  logA B=log base A of B.
EX:  log2 pie = log base 2 of pie.
EX:  log(pie) 2 = log base pie of 2
6)  Prove algebraically, without using logarithm tables or calculators,
that:
        a)  (1/log2 pie) + (1/log5 pie) > 2
        b)  (1/log2 pie) + (1/log(pie) 2)  >2
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From: "Mr. X" <boot-leg@ix.netcom.com>
Subject: hard problems
Date: 1998/02/23
Message-ID: <6ct1fq$jan@dfw-ixnews6.ix.netcom.com>
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