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Solve log52 (x-3) = x

  1. Oct 3, 2013 #1
    1. The problem statement, all variables and given/known data

    log52 (x-3) = x

    2. Relevant equations
    log laws


    3. The attempt at a solution
    One of my friends came up with this equation and asked me to graph it. I haven't seen an equation like this, and do not know how to graph using mapping notation or any other methods. Is it possible to graph this?

    Looking at the equation, I would assume that there can only be one answer to this because

    log52 can only have one answer since this means that 5x = 2

    Can someone please point me in the right direction?

    Thanks :)
     
  2. jcsd
  3. Oct 3, 2013 #2

    DrClaude

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    Staff: Mentor

    You can graph a function, not an equation.
     
  4. Oct 3, 2013 #3
    OH i see thank you
     
  5. Oct 3, 2013 #4

    haruspex

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    Science Advisor
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    True, but maybe what was meant was to graph the function each side of the equation and see where the graphs cross.
    Acnhduy, did you mean (log52) (x-3) or log5(2 (x-3))?
     
  6. Oct 3, 2013 #5
    I was given (log52) (x-3)
     
  7. Oct 3, 2013 #6
    This is not true. The graph of an equation is the set of tuples which make the equation a true statement.
     
  8. Oct 3, 2013 #7
    How do I graph this?
     
  9. Oct 3, 2013 #8
    If you really were given the equation ##\log_5(2)\cdot(x-3)=x##, then you have a linear equation in a single variable. It has one solution (value which makes it true); ##\frac{3\log_5(2)}{\log_5(2)-1}##. The graph is the singleton set ##\{\frac{3\log_5(2)}{\log_5(2)-1}\}##. The picture associated with this graph is rather uninteresting; it's just a point plotted on the real line.
     
  10. Oct 3, 2013 #9
    How did you get to ##\frac{3\log_5(2)}{\log_5(2)-1}## ?

    Is it because you distribute log52 to (x-3)?

    Can you show me the steps, I don't really know what is going on :frown:

    If you distribute it, all I can think of is

    x = xlog52 - 3log52

    the coefficient can become the exponent so

    x = log52x - log523

    if so, since the bases are the same I can divide?

    x= log5 (2x / 8)

    Please show me the steps, this is all I can think of.

    Many thanks.
     
  11. Oct 3, 2013 #10
    Keep in mind that ##\log_5 2## is just a number. For the time being, let's use the letter ##a## in its place. So you have

    ##a(x-3)=x##

    which after distributing the ##a## becomes

    ##ax-3a=x##.

    Grouping the ##x## terms on the left and everything else on the right gives

    ##ax-x=3a##,

    and factoring out ##x##

    ##(a-1)x=3a##.

    Solving for ##x## and plugging ##\log_5 2## back in for ##a## gives

    ##x=\frac{3a}{a-1}=\frac{3\log_5 2}{\log_5 2-1}##.
     
  12. Oct 3, 2013 #11
    Oh! I understand now ! :D

    Would doing this result in the same thing?

    log52 (x-3) = x

    Can I divide both sides by (x-3)

    log52 = x / (x-3)

    And now use the definition of log where

    5x/(x-3) = 2

    I don't know where I am going with this but how do I solve for x in this case, and would it give me the same as
     
  13. Oct 3, 2013 #12
    Not to be flippant, but in order to solve ##5^{\frac{x}{x-3}}=2##, you undo all the things you just did and then do it the way that I showed. There is really no way to do it that involves using any properties of logarithms or exponents.

    If this were an actual problem from a precalc class, it's likely that the problem started with ##5^{\frac{x}{x-3}}=2## and asked you to solve for ##x##.
     
  14. Oct 3, 2013 #13
    Okay then, thanks for the help !
     
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