1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Logs and more
  1. Log-log scales (Replies: 4)

  2. Log questions (Replies: 3)