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Logs and more

  1. Aug 28, 2007 #1
    1. The problem statement, all variables and given/known data
    Solve [tex]2(5^{x+1})=1+\frac{3}{5^{x}}[/tex] giving the answer in the form of [tex]a+\log_{5}b[/tex] where a, b, are a set of integers.
    2. Relevant equations


    #2:[tex]a^{x} / a^{y}=a^{x-y}[/tex]









    #11:[tex]a^{x}=b[/tex] can be put as: [tex] x=\log_{a}b [/tex]

    3. The attempt at a solution
    Here it goes:

    Following #6(backwards), you get


    so [tex]x= -1 + \log_{5}(\frac{3}{(5^{x})(4)}[/tex]

    so wait, lets use #7
    [tex]\log_{5}3 - \log_{5}5^{x}(4)[/tex]

    [tex]\log_{5}3 - \log_{5}5^{x} + \log_{5}4[/tex]

    so [tex]\log_{5}3 - x\log_{5}5 + \log_{5}4 = \log_{5}3 - x + \log_{5}4[/tex] so it equals: [tex]2x=-1 + \log_{5}12[/tex]

    How can I divide the 2?

    How about this..
    does that make [tex]x=\frac{1}{2}+\frac{12}{25}[/tex]? (wait disregard this, it doesnt matter since.. well, it doesnt ask for x)

    **Genneth im trying your way but I sort of dont get it

    **Couple things: I put all the formulas I was given just in case, I may have skipped a few but I don't think any other is relevant. Also, I will do more than one problem in this thread, as I am.. quite stuck.. I'll do my best to minimize the number of problems, if I can solve them hhe.
    Last edited: Aug 28, 2007
  2. jcsd
  3. Aug 28, 2007 #2
    Your application of #6 is backwards...

    Hint: 5^(x+1) == 5*5^x. Try finding 5^x first.
  4. Aug 28, 2007 #3
    I have no idea how that helps...:S?
  5. Aug 28, 2007 #4
    how would you get rid of the denominator on the right side?
  6. Aug 28, 2007 #5
    Well, it equals [tex]log_{5}25[/tex] is the same as 2 (I made it log base 5 of 25 on purpose), but the only way I can get the equation to be in the right format is if it equals '2x', and I dont think I can do that. Wait, would it be -1/2 + 1/2log base 5 of 12?
    Last edited: Aug 28, 2007
  7. Aug 28, 2007 #6


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    Homework Helper

    follow genneth's suggestion... then replace 5^x with a another variable say y... solve for y.
  8. Aug 28, 2007 #7
    multiply both sides by [tex]5^x[/tex] and notice that you have a product which has "the same base"
  9. Aug 28, 2007 #8
    Hmm, I had gotten to that step, so then im supposed to factor
    [tex]10y^{2} - y - 3 = 0? [/tex]

    (I was never really good at factoring, so some help would be appreciated, ill try solving it anyways)

    I think I factoreed correctly:

    y= 3/5

    Now I solve for x (y=5^x) and I use the #11 formula

    So, 5^x=3/5 and 5^x=1/2

    My answer is [tex]x=-1+\log_{5}3[/tex]
    That's for 3/5th's... for 1/2 I got [tex] x=-log_{5}2[/tex] ~ Its not right format than the one they ask, so I only give the first answer?, or both answers?
    Last edited: Aug 28, 2007
  10. Aug 28, 2007 #9
    since one of your solution equals -1/2, what can you do with that solution?
  11. Aug 28, 2007 #10

    I know I can call it extraneous as logs cant be negatives, so we 'delete' or 'exclude' or 'eliminate' that one?
    Last edited: Aug 28, 2007
  12. Aug 28, 2007 #11

    since your calculator is log base 10, you will need to use the log-base-change formula.
  13. Aug 28, 2007 #12
    Psst, read the problem, I think im done since it says
    Give the answer in a log base 5 b
  14. Aug 28, 2007 #13
    sorry :D but come on!!! aren't you curious to know if your answer is correct? i already know whether it is or not ;)

    i'll show you the Proof right now if you want to know.
  15. Aug 28, 2007 #14
    Yup, Im right.

    Haha I guess that's part of the fun in this hehe,

    So I solved for x its -.317393805 (lol)
    So, (2)(5^1-.317393805)=1+3/5^-.317393805
    Eventually you get 6=6 :) (Or 6.0000000005=5.999999996, Im wrong! hahaha)

    **im posting another problem in a couple minutes, im stuck, but ill try and figure it out myself first :P
    Last edited: Aug 28, 2007
  16. Aug 28, 2007 #15
    :-] i would've been stuck on that problem if genneth never posted, any more problems? i'm bored
  17. Aug 28, 2007 #16
    This one:
    Find the exact value of x satisfying the equation

    Give your answer in the form [tex]\frac{\ln a}{\ln b}[/tex] where a, b are a set of integers.

    Formulas im using:
    Same ones as shown above in first problem.

    My solution:
    Ok hows this:


    Now, if I multiply everything my ln would my new step look like this:
    [tex]\ln(3^{x}) + \ln (4^{2x})=\ln(9) + ln (6^{x})[/tex]

    This is where I think I went wrong, is it right?

    **Ok im pretty sure im right, so I continue solving, hmm..Now is right about when I may take hints perhaps?
    Last edited: Aug 28, 2007
  18. Aug 28, 2007 #17
    you're right so far

    now use the power rule and factor out a common term
  19. Aug 28, 2007 #18
    the power rule states that [tex]\log_{a}b^{x}[/tex] can be re-written as [tex]x\log_{a}b[/tex]

    so now just factor out x and solve
  20. Aug 28, 2007 #19
    umm so:
    [tex]x\ln(3) + 2x\ln(4)=2ln(3)+xln(6)[/tex]....
    lemme just stare at that to see what I get

    How am I supposed to factor x out in the right if I dont have an x in 2ln3..
    I get:
    [tex]x(\ln3 + 2\ln4 - \ln6)=2\ln3[/tex]
    So you solve inside the parenthesis
    you eventually get [tex]\ln8^{x}=\ln9[/tex]
    Im getting somewhere.. let me think some more hehe
    Last edited: Aug 28, 2007
  21. Aug 28, 2007 #20
    on a side note, for future problems if you're solving for x

    just bring the x term down, don't bother bringing the other term

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