Math Is Fun Forum

Hi!
I hope you have had a chance to try stefy's suggestion and solved the problem.  If not, then
give it a try before looking at the following proof of the rule involved in solving the problem.

The definition of log is usually introduced via two equalities in two variables, namely,
          x                                                                                 
N = 10    and   x = logN which somewhat obscures the meaning of the logarithm.
   
                                                                                       logN
If we substitute logN for x in the first equality we get  N = 10        ("lay on goodies" to get N.)
which gives us a single equality with a single variable.  And from this single equality we can
basically read the definition of logN:  "logN is the exponent we must apply to 10 to get N."
logN is a NAME for the right exponent that applied to 10 produces N.  So logs are names for
exponents.  And the naming is nice in that it tells us what the base is that it must be applied to
and the result we get when we apply it to that base.  If the base is b then the logarithm is
                                                                                     log N
                                                                                         b
log  N which when applied to base b results in N; that is,  b        = N. 
     b

(In Maple the log is written log[b](N).)

                     logN
Writing N as 10       (by the definition of log) we obtain
                                                                                                          p
     p       logN p        p*logN                                                p        log(N  )
   N  =(10      )  =  10           by a law of exponents.    Also  N   = 10          by the definition of log.
    p                                                                      p
( x   was immediately above but should have been N   left of the equal sign.  That's the edit.)
                                     p
            p*logN        log(N  )
Thus  10          = 10             and since the bases are equal and the quantities are equal then
                                                                             p
the exponents must be equal; that is,  p*logN = log(N  ).  (because log is a 1-1 function.)

                                                p                                                                       p
More generally   p*log N = log (N  )  so the exponent on N in the argument of log (N  )
                                b         b                                                                       b
                                                                                                           p
"comes out front" forming p*log N.   And if b=e then we have p*lnN = ln(N  )
                                             b
                                                                          n  m     nm                       logN  p       p*logN
This is a disguised form of the law of exponents:  (x  )   = x      .  Here it is (10      )   = 10        .

In the original problem the p=3 for x and the p=1/3 for the y.

Then the difference produces a quotient in the argument of the single logarithm.

                                3        1/3             3   1/3
    3lnx-(1/3)lny =  lnx  - ln(y     )  =  ln(x /y     )

                                                                        p
In base 10 the rule can be written:    plogN = logN .
                                                                                             n m      nm
The law of exponents is:  exponentiation produces a product   (N  )   = N     .
                                                    p
Similarly the exponentiation in log(N  )  produces the product P*logN.

The laws of logarithms are disguised forms of the laws of exponents because logarithms ARE
exponents.

  Hasta la vista!  Gotta get some

Hi!

Or using the laws of logarithms the LHS = log [x(x-63)] so  log [x^2-63x] = 2.
                                                                 8                       8
Converting to exponential notation (or making both sides exponents for 8) we have
    2
  8   =  x^2 - 63x  so that  x^2 - 63x - 64 = 0.  And upon factoring the LHS:  (x-64)(x+1) = 0.
Hence x=64 or x=-1.  We can eliminate the -1 if we don't wish to get into complex numbers.

Converting from log notation to exponential notation and visa versa is simple:
           
           log y = z is equivalent to
               x
               
                 z
               x   = y

If we write the base x for both the log and the exponential on the same side (left here) then
the other two quantities ( y and z here) switch sides.
                                                           x
Other examples:  log 32 = x  becomes  2  = 32  (so x=5)
                              2                                               
                                                               x              -2
                          log (1/9) = x  becomes  3  = 1/9 = 3    hence x=-2.
                              3

                            3
                          5   = 125  becomes  log 125 = 3.
                                                            5
and so forth.

logarithms ARE EXPONENTS.  It's just a naming device so we can tell what base they are supposed
to go on and to see what the result should be if we put in on that base.

                    log100                                                                                 
Example:   10           log100 is the exponent we must put on 10 to get 100. 
                                           2
                               Since 10  = 100, log100 = 2.

                   lnx                                                                                  lnx
                 e       lnx is the exponent we must put on e to get x.  Hence e     = x.
                 
                   log N 
                       b
In general  b         = N   so log N is the exponent we must put on b to get N.  (b>0 and b<>1).
                                            b
Have a great day (or night as the case may be)! 

Hi!

The third degree polynomial divided by the fifth degree polynomial can be written as an infinite
series as:

Then just integrate it term by term to get the indefinite integral in an infinite series form.

Starting with the 1/x^2 term the coefficients in the series repeat in blocks of 10 in the pattern
1,-1,1,0,0,-1,1,-1,0,0.  That is what the 1/(x^10n) handles.  So the next 6 terms in the series
would have exponents on the bottom of 12,13,14,17,18,19 and the next 6 would have
22,23,24,27,28,29 for the exponents on the bottom.

The signs in each block of six terms would continue to be 1, -1, 1, -1, 1, -1.

Has anyone got a closed form for the integral?  I'd love to see it.  If x^5+1 is divided by x-1 I get
x(x+1)(x-1)(x^2+1)  +  (x+1)/(x-1)  but I haven't been able to use this to get a closed form.

Good luck with it!

Or maybe the next for 1,2,4 considering it as a list of consecutive positive integers whose spelling is less than five letters long:
1, 2, 4, 5, 6, 9, 10 --- a FINITE sequence. 

Amen to that bobbym!

I hate such questions.  It just shows the ignorance of the test writers.  The sequence could have
been 1,4,16,25,1,4,16,25,1,4,16,25...  or 1,4,16,25,pi,1,4,16,25,pi...  or  any repeating sequence
that begins with 1,4,16,25 among other possibilities like 1,4,16,25,28,30,39,... where one keeps
adding 3 then 12 then 9 repeatedly. 

Also those tests that have 3 (usually) pictures and the test taker is supposed to figure out the fouth
picture (from a list of 4 or more given pictures) are faulty.  Those that don't think the same way
that the test maker does have their answers counted wrong although they may have just as valid
a reason for their choice.

Some of the questions on those tests sometimes show a lack of understanding of the test taker's
creativity and intelligence.

I saw a test once that was testing quality control.  Out of 100 questions 10 of them were messed
up one way or another.  One of the questions was:

    A stock was purchased for $100.  At the end of the 1st month it had lost 50% of its value.  In
the next month it gained 50% of the value it had at the end of the 1st month.  How much was
the stock worth at the end of the 2nd month?
Choices
a)  $100
b)  $75
c)  $50
d)  It's hard to understand the stock market.
e)  $200
Their "correct answer" was b AND d.

Go figure!  How can one pass a test with such questions?
It's easy to understand the stock market:  people buy and sell stocks.
PREDICTING what the stock market is another matter.

Hi!

Of course 1 is a number.  Aren't most people looking out for #1?  That's the most mathematics
that many people want to get involved with.  But then again maybe they just haven't found it yet
so they don't think it exists!

So let's see if we can post a bigger one;             XXXX
                                                                       XXXX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                           XX
                                                                      XXXXXX
                                                                      XXXXXX

                                                 Wow!   That's a BIG one!  
                                         Could LaTex himself even post as BIG a one?

Hi!

Those of you who have access to a good graphing program might try graphing the function
f(x) = 3|x-2|/x and then draw the horizontal line y=4 and see where the graph lies below
or on that line.  I think you may find that any x in (-inf, 0) union [6/7, inf) will have its output
less than or equal to 4. (I graphed it by hand so that could be a "bit shaky".  I'd like to see
the graph of this function using a good graphing program.  Hint!  Hint!)

I'm still a bit confused about what the original question is.  The domain of the function above
is any real number except zero.  The range seems to be the set R-[-3, 0), R being the set of reals.

If we require f(x) <= 4 then this is true for a subset of the domain, namely that mentioned above:
(-inf,0) union [6/7, inf).  Any x in the subset [1, inf) of the domain will certainly satisfy f(x) <= 4.

If we require that f(x) be in [-3, 0) then the domain is restricted to the empty set.
Any restriction on the outputs of f(x) corresponds to a subset of the domain R-{0}.

Solving the original inequality is essentially finding out what subset of the domain of f(x) will make
the inequality true.  To do so I usually consider all possible cases that make a difference in the
expressions replacing the absolute value expressions.  It's a bit tedious, but I tend to get lost if I
don't consider each case.  The union of the sets obtained in each case is typically the final answer.
Also in each case it is the intersection of the assumed condition and its result that gives the set for
that case.

Determining the range of the function is a different matter altogether.

Note:  We could also make the function f(x) = (3|x-2|/x) - 4 and see where it is <= zero.

Absolute value inequalities are tricky.  It is probably best to write an appropriate function f(x)
and graph it to see what the solutions are  keeping in mind that the solution set is a subset of
the domain of the function.

  GOTTA GET SOME 

Hi!

Here's another explanation why division by zero is not possible.  Division is a secondary operation,
not a primary operation; that is, division is not given in the field axioms that establish the real
number system.  Multiplication is given in the axioms (as well as addition, but not subtraction).
Division is then defined  as multiplication by reciprocals.

For example suppose x is a real number whose reciprocal is y.  Then given a number z we define
z/x as z*y.  So to divide by x, x must have a reciprocal.  BUT zero has no reciprocal as stated in
the multiplicative inverse axiom:  For each real non-zero number x, there is a real number y such
that xy=1.  Of course the usual notation for the reciprocal of x is written 1/x.  But this looks like
division, so using 1/x for the reciprocal in the axiom and then later using "/" for division is a bit
confusing and makes the definition via z/x = z*(1/x) look circular.

When all is said and done, the axioms do not allow for a reciprocal of zero, hence division by zero
is a non-issue --- it could never happen since zero has no reciprocal to define division by zero in
terms of.

Having zero in the denominator of a fraction is a similar issue.  The set of fractions based on the
set of whole numbers W={0,1,2,3,...} is defined as

                               F = { p/q | p and q are in W and q is not zero }

If someone asks why we can't have something like 2/0 the answer is simply the definition does
NOT allow zero in the denominator.  It has nothing to do with "division by zero."

Analogously the field axiom for the reals do not allow zero to have a reciprocal.  Hence to write
an expression one would read as "division by zero" simply violates the field axioms.  Trying to
write "a number divided by zero" would be defined as "a number times the reciprocal of zero."
But a "reciprocal of zero" does not exist.

The other explanations of why one can't divide by zero illustrate the problems that would occur
if we did allow zero to have a reciprocal.  As such they provide good reasons from disallowing
zero to have a reciprocal in the first place.

A nice physical example of trying to divide by zero can be seen in the operation of the old
mechanical calculators of yesteryear.  They operated on division as a "repeated subtraction."
If one tried to divide 2 by zero, the calculator would subtract zero from two, add one to the
quotient, and then check to see if there was enough left to subtract zero from it again.  Of course
there was enough left to subtract zero again since 2>0.  So it subtracted zero from two again,
added one to the quotient, and checked to see if there was enough left to subtract zero again.
The old mechanical calculators got stuck in an "infinite loop" subtracting zero over and over
again.  The quotient register looked like an old gas pump register with the dials spinning as
the quotient grew.

A friend of mine in high school rented a mechanical calculator to do lots of arithmetical homework.
On the front in bold letters it said "DO NOT DIVIDE BY ZERO!"  Of course, that was an invitation.
So about half way through his assignment he starts a division by zero.  The calculator was run
on electricity, so it just kept spinning the digits in the quotient.  Quite fun to watch, but tiresome
after a while.  So he punched "clear" and all the other buttons and levers he could find.  Nothing
stopped the process.  Finally he got the idea to pull the plug, and presto! it stopped.

So he thought that he'd better get back to his homework.  He plugged it back in and presto! it
resumed the division calculation.  He did the rest of his assignment by hand.  It appears that
something had to be reset internally to get it to stop the division.  I suspect when he returned
the machine he just set it on the counter and quickly headed for the door!

The old electric mechanical calculators (which were made of metal and weighed about 60 lbs)
demonstrated quite well the "repeated subtraction" algorithm for division.  You could watch it
as it proceeded through a calculation as the dials spun and the carriage shifted.  Slow by today's
standards, but it got the job done (if not dividing by zero!).

Thanks all!

Hi MIF!

Looks like an errorless labor of love to me! 

How does this look? :0)
                        i*180                    i*(180/n)                                  i0
(-1)^(1/n) = (1*e       )^(1/n) = 1*e              so this approaches 1*e   = 1 as n goes to infinity.

(The angles are in degrees.)