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    Solve: Log2x + Log4x = 2

    I did 1/Logx2+ 1/Logx4 = 2, but am now stuck on how to proceed.
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    change of base rule?
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    (Original post by sulaimanali)
    change of base rule?
    What do you mean?
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    (Original post by Desultory)
    What do you mean?
    Have you learnt the change of base rule for logarithms?
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    (Original post by RDKGames)
    Have you learnt the change of base rule for logarithms?
    well clearly they haven't
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    (Original post by ZiggyStardust_)
    well clearly they haven't
    ...or they forgotten that they have.
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    it is a thing:

    logax ≡ { logbx } / logba

    for instance

    log28 = { log168 } / log162

    3 = (3/4) / (1/4)
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    (Original post by the bear)
    it is a thing:

    logax ≡ { logbx } / logba

    for instance

    log28 = { log168 } / log162

    3 = (3/4) / (1/4)
    I know that, but I still dont know how to solve it?
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    (Original post by Desultory)
    I know that, but I still dont know how to solve it?
    Post your attempt. I recommend switching \log_4(x) into base 2.
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    (Original post by RDKGames)
    Have you learnt the change of base rule for logarithms?
    Ihave learnt them all
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    (Original post by RDKGames)
    Post your attempt. I recommend switching \log_4(x) into base 2.
    I did that. I got up to:
    Log24 + 1 = 2Logx4
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    (Original post by Desultory)
    I did that. I got up to:
    Log24 + 1 = 2Logx4
    OK... so then what is the value of \log_2(4)? Get the form a=\log_x(4) then exponentiate with an appropriate base and solve from there.

    Alternatively;
    Spoiler:
    Show

    You should know that \log_4(x)=\frac{\log_2(x)}{\log_  2(4)} so your equation becomes \log_2(x)+\frac{1}{2}\log_2(x)=2 which is easier to solve
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    (Original post by RDKGames)
    OK... so then what is the value of \log_2(4)? Get the form a=\log_x(4) then exponentiate with an appropriate base and solve from there.

    Alternatively;
    Spoiler:
    Show


    You should know that \log_4(x)=\frac{\log_2(x)}{\log_  2(4)} so your equation becomes \log_2(x)+\frac{1}{2}\log_2(x)=2 which is easier to solve

    Ok got your way thanks
 
 
 
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