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    Solve log base 2a 4a squared
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    (Original post by Niki5)
    Solve log base 2a 4a squared
    Is that \log_{2a} (4a)^2?

    Or \log_{2a} (4a^2)?

    If the latter, remember that \log_x y^n = n \log_x y
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    It is not (4a) squared but it is 4a squared
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    (Original post by joostan)
    Is that \log_{2a} (4a)^2?

    Or \log_{2a} (4a^2)?

    If the latter, remember that \log_x y^n = n \log_x y
    yes it is the second thing you wrote joostan
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    (Original post by Niki5)
    yes it is the second thing you wrote joostan
    Read the bottom of my first post
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    Thank you for your help. I got it.
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    (Original post by Niki5)
    Thank you for your help. I got it.
    Good
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    wait a minute
    /log-{2a}{4a^2}
    is equal to
    2log-{2a}4a
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    Do i ignore the 2 infront?
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    (Original post by Niki5)
    wait a minute
    /log-{2a}{4a^2}
    is equal to
    2log-{2a}4a
    No.
    \ (2a)^2 = 4a^2

\Rightarrow \log_{2a} (4a^2) = . . .
    Finish off yourself
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    (Original post by joostan)
    No.
    \ (2a)^2 = 4a^2

\Rightarrow \log_{2a} (4a^2) = . . .
    Finish off yourself
    To me it looks like the answer is 2 but i don't understand how you came up with (2a)2=4a2
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    (Original post by Niki5)
    To me it looks like the answer is 2 but i don't understand how you came up with (2a)2=4a2
    Correct
    Just rules of indices . . .
    So \ (2a)^2 = 2^2 \times a^2 = 4a^2
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    (Original post by Niki5)
    To me it looks like the answer is 2 but i don't understand how you came up with (2a)2=4a2
    It is ok I understand it
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    (Original post by Niki5)
    It is ok I understand it
    good
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    How do you solve this one?
    2loga-3-log2c
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    (Original post by Niki5)
    How do you solve this one?
    2loga-3-log2c
    Is that \ 2 \log_{10} (a) - 3 - log_{10} (2c)?
    You can't solve it, it's not an equality.
    To express in a single logarithm: Use \log_x (y) - log_x (z) = log_x (\frac{y}{z})
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    (Original post by joostan)
    Is that \ 2 \log_{10} (a) - 3 - log_{10} (2c)?
    You can't solve it, it's not an equality.
    To express in a single logarithm: Use \log_x (y) - log_x (z) = log_x (\frac{y}{z})
    well they wrote lg instead of log
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    Leave the last one
    Can you help me with this one then?
    2 to the power of 2x minus 2 to the power of x minus 6
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    (Original post by Niki5)
    Leave the last one
    Can you help me with this one then?
    2 to the power of 2x minus 2 to the power of x minus 6
    That's another expression. Do you mean 2^{2x} - 2^x - 6=0?

    You can factorise it like a quadratic.
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    (Original post by Damask-)
    That's another expression. Do you mean 2^{2x} - 2^x - 6=0?

    You can factorise it like a quadratic.
    True say
 
 
 
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