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    (Given that log (base 3) x=2 determine the value of y for which log (base 3) (y+4)=2.
    Please help. I do not know what to do when there are 2 values in that position.
    I actually figured it out. But i'd like to know why the textbook only says it's Y=12, not y=-3.. y=-3 seems to work as well. So is the textbook answer wrong?
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    (Original post by MathMeister)
    (Given that log (base 3) x=2 determine the value of y for which log (base 3) (y+4)=2.
    Please help. I do not know what to do when there are 2 values in that position.
    I actually figured it out. But i'd like to know why the textbook only says it's Y=12, not y=-3.. y=-3 seems to work as well. So is the textbook answer wrong?
    Not sure what the x bit is about

    \log_3(y+4) = 2

    Means

    y + 4 = 3^2

    So I am not sure where your or the books answers come from


    Can you type in the exact question
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    (Original post by TenOfThem)
    Not sure what the x bit is about

    \log_3(y+4) = 2

    Means

    y + 4 = 3^2

    So I am not sure where your or the books answers come from


    Can you type in the exact question
    Ignore the x. I thought the question carried on from the first one- but it seems it didn't.
    Uhhm.. Yes, you are right that \log_3(y+4) = 2 Means y + 4 = 3^2, however since the logs in the equation are being subtracted from each other, you divide them and you get log (base 3) ([y^2)/(y+4)] =2 ... Then I solved it.
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    (Original post by MathMeister)
    Ignore the x. I thought the question carried on from the first one- but it seems it didn't.
    Uhhm.. Yes, you are right that \log_3(y+4) = 2 Means y + 4 = 3^2, however since the logs in the equation are being subtracted from each other, you divide them and you get log (base 3) ([y^2)/(y+4)] =2 ... Then I solved it.
    mm you sure that is the question? Where in the book is it from?
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    (Original post by Super199)
    mm you sure that is the question? Where in the book is it from?
    Yes, Its on page 49 of the C2 edexcel modular book, chapter 3.6 Mixed exersise 3G question 4b
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    (Original post by MathMeister)
    Yes, Its on page 49 of the C2 edexcel modular book, chapter 3.6 Mixed exersise 3G question 4b
    The question is:

    2log_3y-log_3(y+4)=2 which isn't what you posted

    Use power rule
    Then the subtraction one
    anti-log

    You should get a quadratic, solve for y. One of your answers will be invalid because you cannot log a negative.
    Post your workings if you get stuck
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    (Original post by MathMeister)
    Ignore the x. I thought the question carried on from the first one- but it seems it didn't.
    Uhhm.. Yes, you are right that \log_3(y+4) = 2 Means y + 4 = 3^2,



    however since the logs in the equation are being subtracted from each other, you divide them and you get log (base 3) ([y^2)/(y+4)] =2
    ... Then I solved it.
    I do not mean to be rude but you really need to ask the question that you want answering
 
 
 
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