



Example:



What do
we do with this one?






log(X + 3)
+ log(2X  5) = 2






Remember
the rule of logs that said: 





log(a × b)
= log a + log b






We're
going to use that rule, 


but going
in the opposite direction ... 





log a +
log b = log(a × b)






So: 


log(X + 3)
+ log(2X  5) = 2



log((X +
3)(2X  5)) = 2






Multiply it
out ...



log(2X^{2}
+ X  15) = 2






OK, now
what? 





Well
remember that when we write log we really mean log 10 


so let's
write the problem like that ... 





log_{10}(2X^{2}
+ X  15) = 2






Now what? 





When in
doubt, read it out! 


Read the
meaning of the statement 











Or,
written without logs ... 











And while
we don't have any tricks to help us solve 


something
like log_{10}(2X ^{2} + X  15) = 2 


we can
solve: 


10^{2}
= 2X^{2} + X  15






Here goes
... 








OK, where
did we leave that old quadratic formula ... 





Here it
is: 


A =
2 B = 1 C =  115












Now we
take our two answers back to the original problem 


and see
if they make sense ... 





log(X + 3)
+ log(2X  5) = 2



X = 7.34



log(7.34 +
3) + log(2(7.34)  5) = 2



log(10.34)
+ log(9.68) = 2






Punch
these two up on your calculator and you get ... 





1.01 + .99
= 2






It
checks! 





Now try
the other one ... 





log(X + 3)
+ log(2X  5) = 2



X = 7.34



log(7.34
+ 3) + log(2(7.34)  5) = 2



log(4.83)
+ log(20.66) = 2






AND HERE
WE HAVE A BIG PROBLEM 





Do you
remember what logs mean? 











And
there's no exponent in the world that we can raise 10 to 


to get
minus anything. 


The only
thing we can get is positive numbers. 


Even
negative exponents give us positive numbers. 


That
means if you have something like: 





log_{10}X






X MUST BE
POSITIVE






So X =
7.83 does NOT check. It is not a solution to the problem. 





The
numbers that X can be are called it's DOMAIN 


So for
stuff like Log X (and ln X too), 


the
domain is all real numbers greater than zero. 





If we had
... 


Y
= ln(X1)






The (X1)
part all together must be positive. 


That
means X  1 > 0 





So: 








The
"domain" of X in Y = ln(X1) is X
>1 "all real
numbers greater than 1" 





copyright 2005 Bruce Kirkpatrick 
