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How to solve the equation 3^2x+1 = 3^x + 24, do I need logarithms

  1. Apr 8, 2004 #1
    Greetings all, I have a slight problem solving this equation ..... 3^2x+1 = 3^x + 24

    I know I can solve it with logs but I'm just not sure where to begin. I have tried taking logarithms both sides but it didn't work :confused:

    Perhaps someone could give me a clue :biggrin:

    thanks
     
  2. jcsd
  3. Apr 8, 2004 #2
    Can you show the work you've done so far?

    cookiemonster
     
  4. Apr 8, 2004 #3
    Sure...

    3^2x+1 = 3^x + 24

    (2x+1)lg3 = xlg3 + lg24

    (2x+1)lg3/lg3 = xlg3/lg3 + lg24/lg3

    2x+1 = x + 2.89

    x = 1.89

    This is obviously wrong.
     
  5. Apr 8, 2004 #4
    Be careful with log (3^x + 24).

    In general, log(a + b) does not equal log(a) + log(b).

    And you need to be careful with your notation. I first read the problem as 3^(2x) + 1 = 3^x + 24. Those darned parentheses!

    cookiemonster
     
  6. Apr 8, 2004 #5
    32x+1 = 3x + 24

    Can you show that 32x+1 = 3*32x? Can you show that 3*(32x) = 3*(3x)2? Now what kind of equation do you get if you let t = 3x?
     
  7. Apr 8, 2004 #6
    Thank you very much for your help.
     
  8. Apr 10, 2004 #7
    These logarithms is not my good friend unfortunitaly, it is either that or I'm missing a fundamental point.

    [itex] log_3x - 2log_x3 = 1 [/itex]

    I don't understand how to solve this, I know the rules of logarithms but they don't seem to be much help here. Can anyone help me please.

    Thanks
     
  9. Apr 10, 2004 #8
    You might find this useful

    [tex]\log_b a = \frac{\ln a}{\ln b}[/tex]

    cookiemonster
     
  10. Apr 10, 2004 #9
    More generally:

    [tex]\log_b a = \frac{\log_c a}{\log_c b}[/tex]

    Where c is whatever number you want (cookiemonster chose c = e, resulting in natural logs). For the problem above you might want to pick c = 3.
     
  11. Apr 11, 2004 #10
    It worked !! Thanks for the help yet again. One question though, how did you come to the conclusion that I should take the base as 3?
     
  12. Apr 11, 2004 #11
    You started with this:

    [itex] log_3x - 2log_x3 = 1 [/itex]

    Using ln you would have gotten:

    [itex] \frac{\ln x}{\ln 3} - 2\frac{\ln 3}{\ln x} = 1 [/itex]

    And just get stuck with ln(3). However, using 3 as the base you can get:

    [itex] \frac{\log_3 x}{\log_3 3} - 2\frac{\log_3 3}{\log_3 x} = 1 [/itex]

    And of course log3(3) = 1 and it's much easier to handle:

    [itex]\log_3 x - \frac{2}{\log_3 x} = 1 [/itex]

    Obviously you would still get the same answer with both methods, I just thought it would be slightly easier to solve when using base 3. :smile:
     
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