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Two logarithm problems

  1. Jul 26, 2004 #1
    I've just introduced myself to logarithms and have done most of the questions, but am having trouble with one or two of them:

    Q1: Find values of x for which:
    Log(to base 3)x - 2log(to base x)3 = 1.

    I have no idea where to start on this question.


    Q2: Solve:
    25^x = 5^(x+1) -6.
    On this question, i know it can be simplified to give:
    25^x = 5(5^x) - 6
    However, if i let y=5^x, what would 25^x be? It's not 5y, as 5(5^x) is 5y.

    Thanks for any help.
     
    Last edited: Jul 26, 2004
  2. jcsd
  3. Jul 26, 2004 #2
    Q1: Use the change of base rule to convert [itex] \log_x{3}[/itex] to some base 3 (or both to base 10 if you prefer) log.

    Do you know your rules of logs? Do you know what to do with a coefficient in front of a log?

    Q2: Is it supposed to be written as

    [tex]25^x = 5^{(x+1)} -6[/tex]

    or

    [tex]25^x = 5^{(x+1) -6}[/tex]
     
  4. Jul 26, 2004 #3
    Q2 is the first one. The -6 is completely seperate and not part of the logarithm.
     
  5. Jul 26, 2004 #4
    Q1: Log(base3)x - 2log(basex)3 = 1
    Log(base3)x - log(basex)9 = 1.
    (Lgx/lg3) - (lg9/lgx) = 1 ?

    ( (lgx)^2 - (lg9)(lg3) )/(lg3)(lgx) = 1 ?
    (lgx)^2 - (lg9)(lg3) = (lg3)(lgx)
    I'm not sure how i would evaluate that.
     
    Last edited: Jul 26, 2004
  6. Jul 26, 2004 #5
    okay, well its been a little while since i've done logs, but i'm pretty sure this is how it goes....

    remember the properties of logarithms:
    log(x)^y = ylogx
    is one of them, i remember, and i'm sure there's one regarding multiplication, but its not coming ot me at the moment.

    good luck
     
  7. Jul 26, 2004 #6
    That looks OK.

    Let u = lg x. Then the above becomes

    u2 - u lg(3) - lg(9)lg(3) = 0

    which you can solve using the quadratic equation.
     
  8. Jul 26, 2004 #7
    Note that 25x = (52)x = (5x)2. Then let y = 5x as you suggest and you end up with a quadratic which you should know how to solve.
     
  9. Jul 27, 2004 #8
    Thanks for the help. I managed to solve the second one thanks tot he quadratic equation. For some reason i thought (5^x)^2 would give 25^2x. (In a similar manner to how (5x)(5x) gives 25x^2)

    I'll do the other one now, thanks.

    Say you have Log(baseA)B. When converting to a different base this can become (log(basez)(B)) / (log(basez)(A))

    Where base Z is any base you like, such as e or 10 ?
    Thanks.
     
  10. Jul 27, 2004 #9

    AKG

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    Yes, if you mean what I think you mean:

    [tex]\log _A B = \frac{\log _z B}{\log _z A}[/tex]
     
  11. Jul 27, 2004 #10
    Thats the one, thanks.
    Just proof, a little integration i haven't done yet and numerical methods before i've finished this book >_<.
     
  12. Jul 27, 2004 #11
    What I do whenever I forget a particular equation (like all those crazy trig. identities) is derive them. For the equation in question,
    [tex]\begin{align*}
    x &= x \\
    a^\log_a{x} &= b^\log_b{x} \\
    (b^{\log_b{a}})^\log_a{x} &= b^\log_b{x} \\
    b^{\log_b{a}\log_a{x}} &= b^\log_b{x} \\
    \log_b{a}\log_a{x} &= \log_b{x} \\
    \log_a{x} &= \frac{\log_b{x}}{\log_b{a}}
    \end{align}[/tex]

    Easy as eating cake.
     
  13. Jul 27, 2004 #12

    Gokul43201

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    For the first problem, there is no need to use the change of base rule. Just call [tex]~~log_3 ~x = y ~~[/tex] and the rest follows simply enough.
     
    Last edited: Jul 27, 2004
  14. Jul 28, 2004 #13
    I got it and finished that chapter, thanks for the help.
    Deriving does help sometimes, if you can do it for that particular formula easily enough. An example is the double angle to single angle formulae - which are just based from sin(a+b) etc.
    Just a quick, small question on some integration - volumes of revolution. On a 360 degrees (2pi) revolution you would multiply the final integrated product by pi. On a 180 degrees revolution you would multiply the final integrated product by pi/2 ?
     
  15. Jul 28, 2004 #14

    AKG

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    Not exactly sure what you're saying. If you have some function revolved [itex]\theta[/itex] about the x axis from a to b, then the volume is:

    [tex]\int _a ^b \frac{\theta}{2} [f(x)]^2 dx[/tex]

    Why is it like this?

    [tex]\frac{\theta}{2} [f(x)]^2 = \frac{\theta}{2\pi}\pi [f(x)]^2[/tex]

    [itex]\pi [f(x)]^2[/itex] is the area of a circle of radius f(x), but since you don't have a full circle, only a sector of a circle, then the area of the sector should be smaller than the area of the full circle by the ratio that the angle is smaller than the full angle (2[itex]\pi[/itex]). That's where the [itex]\frac{\theta}{2\pi}[/itex] comes in. So that gives you the cross sectional area of your little cylinder, then you multiply by the width, [itex]dx[/itex], and then add the volumes of all those cylinders from x=a to x=b.

    So, if by "final integrated product" you meant:

    [tex]\int _a ^b [f(x)]^2 dx[/tex]

    then yes, you're right, but I have no idea why "final integrated product" would mean that. Well, I'm not sure why you use the word "product", and what you mean by "final." I'm not sure why you'd start by doing that integration first, then multiplying by the angle, it seems you'd put all that in the integration.
     
    Last edited: Jul 28, 2004
  16. Jul 28, 2004 #15
    That's what i meant. Now i can work it out for any angle. Thanks.
    Sorry for the bad wording, i'll have to look up how to use the mathematical notation sometime.
    In the book I am working out of Pi is usually left to the left of the integration sign and applied after the limited have been applied but there's really no difference.
     
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