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Homework Help: 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

    #1:[tex](a^{x})(a^{y})=a^{x+y}[/tex]

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

    #3:[tex](a^{x})^{y}=a^{xy}[/tex]

    #4:[tex]a^{0}=1[/tex]

    #5:[tex]a^{1}=a[/tex]

    #6:[tex]\log_{a}xy=\log_{a}x+\log_{a}y[/tex]

    #7:[tex]\log_{a}(\frac{x}{y})=\log_{a}x-\log_{a}y[/tex]

    #8:[tex]\log_{a}x^{y}=y\log_{a}x[/tex]

    #9:[tex]\log_{a}1=0[/tex]

    #10:[tex]\log_{a}a=1[/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:
    [tex]2(5^{x+1})=1+\frac{3}{5^{x}}[/tex]
    So
    [tex]5^{x+1}=1/2+\frac{3}{(5^{x})(2)}[/tex]

    Following #6(backwards), you get
    [tex]x+1=\log_{5}(\frac{1}{2})(\frac{3}{(5^{x})(2)}[/tex]

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

    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..
    [tex]x=-1/2+\frac{\log_{5}12}{\log_{5}25}[/tex]
    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

    learningphysics

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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:
    (5y-3)(2y+1)

    y= 3/5
    y=-1/2

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

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

    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
    hmm

    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
    yesss

    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
    [tex](3^{x})(4^{2x+1})=6^{x+2}[/tex]

    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:

    [tex](3^{x})(4^{2x+1})=6^{x+2}[/tex]
    So
    [tex](3^{x})(4^{2x})(4)=(36)(6^{x})[/tex]
    So
    [tex](3^{x})(4^{2x})=(9)(6^{x})[/tex]

    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

    [tex]x\ln{3}+x\ln{16}=\ln{9}+x\ln{6}[/tex]
     
  22. Aug 28, 2007 #21
    Oh, right, yeah I guess its a bit simpler :p

    umm.. [tex]8^{x}=9[/tex]
    so [tex]x=log_{8}9[/tex]
    change base to e
    [tex]x=\frac{\ln9}{\ln8}[/tex]

    YEA!
     
    Last edited: Aug 28, 2007
  23. Aug 28, 2007 #22
    you're correct, but you're solving for x so you don't need to put it back in standard form

    when you solve for x such as a problem like [tex]5x=4[/tex] you just divide, right?

    it's the same for logs.
     
  24. Aug 28, 2007 #23
    :) I guess so, but now I can just go ln9/ln8, which equals 1.057
     
  25. Aug 28, 2007 #24
    when verifying your answer, you want don't want to round off

    this is how i check my answer on my TI with big numbers (sure i could type it all out again but screw that)

    (answer) press STO-> then ALPHA then choose a letter, then just press ALPHA (letter) whenever u plug in

    idk if u already do that, but just a tip, it becomes really useful in Chemistry
     
  26. Aug 28, 2007 #25
    ..I just started using graphic calcs.. i usually stick to my faithful casio fx-350ms
    but my graphic one is fx-9750G plus.. where the hell is the STO button? lol
     
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