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2E: Laws of logarithms, (3) (c) (i) Watch

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    Help with (c) (i) please :puppyeyes:
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    (Original post by IFoundWonderland)
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    Help with (c) (i) please :puppyeyes:
    Darling, dear, just change the base of the logarithm. Not that hard..

    Ultimately, you get a quadratic using y-substitution which you can solve quite simply. What course are you doing? This is a joke.
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    (Original post by Bath_Student)
    Darling, dear, just change the base of the logarithm. Not that hard..
    I haven't done this in about a year and am re self teaching for a resit from a *****y textbook. Idek what that means :bawling:
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    (Original post by IFoundWonderland)
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    Help with (c) (i) please :puppyeyes:
    I wasn't going to help at first, but then I saw the puppyeyes ...

    If \log_4 x = y then

    4^y = x \Rightarrow (2^2)^y =x \Rightarrow 2^{2y}=x \Rightarrow \log_2 x = 2y = 2 \log_4 x
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    (Original post by IFoundWonderland)
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    Help with (c) (i) please :puppyeyes:
    try these on your calculator and see if it gives you some clues

    log base 3 of 3

    log base 1.4 of 1.4
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    (Original post by IFoundWonderland)
    I haven't done this in about a year and am re self teaching for a resit from a *****y textbook. Idek what that means :bawling:
    Alright!

    So, change log(4) x into base 2, gives you log(2) x / log(2) 4. It's a rule, learn it.
    log(2) 4 is 2. So, you get that log(4) x = log(2)x / 2. Multiply through by 2 to remove the 2 in the denominator. Now, then, the other log(2) x is doubled (as you multiplied through by 2), and a law of logs states that 2*log(2) x = log(2) x^2.

    Now you have a quadratic, let y = log(2) x. Got it from there?!
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    (Original post by Bath_Student)
    Darling, dear, just change the base of the logarithm. Not that hard..

    Ultimately, you get a quadratic using y-substitution which you can solve quite simply. What course are you doing? This is a joke.
    that is a very unhelpful comment.
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    (Original post by atsruser)
    I wasn't going to help at first, but then I saw the puppyeyes ...

    If \log_4 x = y then

    4^y = x \Rightarrow (2^2)^y =x \Rightarrow 2^{2y}=x \Rightarrow \log_2 x = 2y = 2 \log_4 x
    Are you drunk?
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    (Original post by Bath_Student)
    Darling, dear, just change the base of the logarithm. Not that hard..

    Ultimately, you get a quadratic using y-substitution which you can solve quite simply. What course are you doing? This is a joke.
    I do the IB.

    (Original post by atsruser)
    I wasn't going to help at first, but then I saw the puppyeyes ...

    If \log_4 x = y then

    4^y = x \Rightarrow (2^2)^y =x \Rightarrow 2^{2y}=x \Rightarrow \log_2 x = 2y = 2 \log_4 x
    (Original post by maggiehodgson)
    try these on your calculator and see if it gives you some clues

    log base 3 of 3

    log base 1.4 of 1.4
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    The correct answer is 1/4. Also it is a non calculator question *sigh*
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    (Original post by IFoundWonderland)





    The correct answer is 1/4. Also it is a non calculator question *sigh*
    AHHH!!!!!!!!!!!

    I reiterate:

    Alright!

    So, change log(4) x into base 2, gives you log(2) x / log(2) 4. It's a rule, learn it.
    log(2) 4 is 2. So, you get that log(4) x = log(2)x / 2. Multiply through by 2 to remove the 2 in the denominator. Now, then, the other log(2) x is doubled (as you multiplied through by 2), and a law of logs states that 2*log(2) x = log(2) x^2.

    Now you have a quadratic, let y = log(2) x. Got it from there?!
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    (Original post by Bath_Student)
    Darling, dear, just change the base of the logarithm. Not that hard..

    Ultimately, you get a quadratic using y-substitution which you can solve quite simply. What course are you doing? This is a joke.
    no need to be so patronising an unpleasant :erm:

    good luck wonderland :lovehug: xxx
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    (Original post by IFoundWonderland)
    I do the IB.
    In the UK?
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    (Original post by IFoundWonderland)
    I do the IB.




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    The correct answer is 1/4. Also it is a non calculator question *sigh*
    Well, I was just trying to get you to experiment to see that log to base a of a is 1.

    Form that you should be able to do something with line 1.
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    (Original post by Bath_Student)
    When women ransack the mathematics thread!
    how evil of them :afraid:
    tbf i don't know anything about maths so i'll leave lol just came cos wonderland is here
    these ppl could help you tho wonderland lovely :lovehug:
    jamestg
    Student403
    TeeEm
    SeanFM
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    (Original post by CoolCavy)
    how evil of them :afraid:
    tbf i don't know anything about maths so i'll leave lol just came cos wonderland is here
    these ppl could help you tho wonderland lovely :lovehug:
    Damnit I answered her question. If her IQ is larger then 3, then she'll be fine.
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    (Original post by Bath_Student)
    AHHH!!!!!!!!!!!

    I reiterate:

    Alright!

    So, change log(4) x into base 2, gives you log(2) x / log(2) 4. It's a rule, learn it.
    log(2) 4 is 2. So, you get that log(4) x = log(2)x / 2. Multiply through by 2 to remove the 2 in the denominator. Now, then, the other log(2) x is doubled (as you multiplied through by 2), and a law of logs states that 2*log(2) x = log(2) x^2.

    Now you have a quadratic, let y = log(2) x. Got it from there?!
    Um

    So when the x is squared its the equivalent of the whole log being squared?

    I didn't realise that :/

    Christ I'm so confused rn and my notes look so messy
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    (Original post by the bear)
    that is a very unhelpful comment.
    I already spent my bear rep™ today it seems… PRSOM.
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    (Original post by IFoundWonderland)
    Um

    So when the x is squared its the equivalent of the whole log being squared?

    I didn't realise that :/

    Christ I'm so confused rn and my notes look so messy
    n*log a = log a^n. You must have seen that before?

    Therefore, 2 log(2) x = log(2) x^2 .
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    (Original post by CoolCavy)
    how evil of them :afraid:
    tbf i don't know anything about maths so i'll leave lol just came cos wonderland is here
    these ppl could help you tho wonderland lovely :lovehug:
    jamestg
    Student403
    TeeEm
    SeanFM
    :wavey:

    :mad:
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    (Original post by CoolCavy)
    how evil of them :afraid:
    tbf i don't know anything about maths so i'll leave lol just came cos wonderland is here
    these ppl could help you tho wonderland lovely :lovehug:
    jamestg
    Student403
    TeeEm
    SeanFM
    teaching at present
 
 
 
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