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Math Help - Proving Logs

  1. #1
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    Proving Logs

    Prove:
    [attached] ...
    I have no idea on how to even approach this question, since i only know how to prove ones with the same base.
    please help as soon as possible.
    thanks.


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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by incognito_301 View Post
    Prove:
    [attached] ...
    I have no idea on how to even approach this question, since i only know how to prove ones with the same base.
    please help as soon as possible.
    thanks.


    Hint: Here are the formulas you need to know:

    (1) The change of base formula: \log_a b = \frac {\log_c b}{\log_c a}

    (2) Cor. of (1): \log_a b = \frac 1{\log_b a} ....(by taking c = b in part (1))

    a and b are general terms here, not the a and b in your problem.

    Apply the above formulas in the order they are given and you will have your proof. just to give you a direction (the manipulation can work in either) start with the left hand side of your equation
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  3. #3
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    ok so I got this:
    [attached]
    now what?
    cant think of a single thing to do now
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by incognito_301 View Post
    ok so I got this:
    [attached]
    now what?
    cant think of a single thing to do now
    you are not listening to what i am saying. i said apply the rules in the order i gave you. clearly you applied the second one first.

    look at where you are starting and where you want to get. x and y are not the base of any log here, why did you make them the bases? you always want the bases to be a or b. now you want to start with an expression that has \log_a x and \log_b y and end up with an expression that has \log_a y and \log_b x

    try again
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  5. #5
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    Hello, incognito_301!

    I must assume that you know the Base-Change Formula:

    . . . \log_Q(P) \:=\:\frac{\log(P)}{\log(Q)}


    Prove: . \log_a(x)\cdot\log_b(y) \:=\:\log_a(y)\cdot\log_b(x)

    The left side is:

    . . \log_a(x)\cdot\log_b(y) \:=\:\frac{\log(x)}{\log(a)}\cdot\frac{\log(y)}{\l  og(b)} \;=\;\frac{\log(y)}{\log(a)}\cdot\frac{\log(x)}{\l  og(b)} . = \;\log_a(y)\cdot\log_b(x)

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  6. #6
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    wow, this was so easy, i'm almost ashamed that i asked it in the first place. lol
    this is truly the best help-forum that i've come across: if you do your best to solve a question, people will help you. amazing forum. thank you.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, incognito_301!

    I must assume that you know the Base-Change Formula:

    . . . \log_Q(P) \:=\:\frac{\log(P)}{\log(Q)}



    The left side is:

    . . \log_a(x)\cdot\log_b(y) \:=\:\frac{\log(x)}{\log(a)}\cdot\frac{\log(y)}{\l  og(b)} \;=\;\frac{\log(y)}{\log(a)}\cdot\frac{\log(x)}{\l  og(b)} . = \;\log_a(y)\cdot\log_b(x)

    Nice solution.

    This is the one i was thinking of

    \log_a x \cdot \log_b y = \frac {\log_b x}{\log_b a} \cdot \frac {\log_a y}{\log_a b} ..........by formula (1) in post #2

    ................... = \log_b x \cdot \log_a y

    (since \log_b a \cdot \log_a b = 1 by formula (2) in post #2)
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