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Solving logarithms

  1. Feb 16, 2006 #1
    I cant seem to find where im going wrong on this Question. it asks to solve:

    log_3 (2x+3) - log_3(X+1) = 2 where _3 is the base of 3 for log

    so far i moved the log_3(x+1) to the right side of equal sign. then i moved the 2 up as an exponent:

    log_3(2x+3) = log_3(x+1)^2 then i cancelled the logs out and moved the 2 back down so its :

    2x+3 = 2(x+1)

    then distributed the 2 to the x+1 , but then the x's cancel.

    i think that maybe i should have just left it as (x+1)^2 ? is that where im going wrong? or do i actually have to divide 2x+3 by x+1 instead of moving the log_3(x+1) over?
     
  2. jcsd
  3. Feb 16, 2006 #2
    Logarithms have the property that log(a)-log(b)=log(a/b) if they have the same base. So you do this to combine the logs, next you undo the logs by raising the base of the log to the power of both each side i.e. if log(a/b)=c where c is a constant then a/b=k^c where k is the base of the log. After this it is a matter of algebra.
     
  4. Feb 16, 2006 #3
    okay, thats much clearer, thnx.
    now I'm a bit confused on how to graph logs.
    log_5(y+2)=x+1

    What I did, changed the whole equation into Exponential Form. I then graphed that and then switched the X and Y coordinates.

    Is that the correct procedure? Do I only interchange the X and Y coordinates once? <not once and the beginning then again after graphing?>
     
  5. Feb 17, 2006 #4
    When you switch the y and the x cooridinates to find a solution by graphing, solve for y. Or in other words, find the inverse of the function, and before you do that find out if the function is one-to-one.
     
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