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    So these are the first two questions on my homework and I was just wondering if anyone could tell me whether or not I have done them right or if theres anything else I need to do? I've literally forgot everything about logs

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    teachers are getting lazy ...
    call that a homework ...
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    (Original post by jessyjellytot14)
    So these are the first two questions on my homework and I was just wondering if anyone could tell me whether or not I have done them right or if theres anything else I need to do? I've literally forgot everything about logs

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    One log rule that you're missing is that \log_a a = 1
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    (Original post by TeeEm)
    teachers are getting lazy ...
    call that a homework ...
    hahaha I wish that was just my homework These are just the first two questions but I wanted to make sure I was doing it right
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    (Original post by jessyjellytot14)
    hahaha I wish that was just my homework These are just the first two questions but I wanted to make sure I was doing it right
    I was gonna say ...
    makes more sense now.
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    (Original post by Zacken)
    One log rule that you're missing is that \log_a a = 1
    Oh right, so would the answer to the first one just be 1 + 1/2 which would equal 3/2 ?
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    (Original post by jessyjellytot14)
    hahaha I wish that was just my homework These are just the first two questions but I wanted to make sure I was doing it right
    Also, you're not quite getting the point of the second question. You can either note:

    \displaystyle \begin{equation*}\log_2 \frac{1}{8} = \log_2 2^{-3} = -3 \log_2 2 = \cdots \end{equation*}

    Or, with your method: \log_2 1 - \log_2 8 = \log_2 1 - \log_2 2^3 = \cdots

    Note that \log_{\text{anything}} 1 = 0
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    (Original post by jessyjellytot14)
    Oh right, so would the answer to the first one just be 1 + 1/2 which would equal 3/2 ?
    Yep! Another way of seeing that is that a^2 \times \sqrt{a} = a^2 \times a^{1/2} = a^{3/2}, so taking the logarithm of that is \log_a a^{3/2} = \frac{3}{2}. But your method is just as fine, I'm just providing a different perspective.
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    for the second question log small 2 (big 8) means 2 to the power something = 8.
    and log small 2 (big 1) means 2 to the power something = 1
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    (Original post by Zacken)
    Also, you're not quite getting the point of the second question. You can either note:

    \displaystyle \begin{equation*}\log_2 \frac{1}{8} = \log_2 2^{-3} = -3 \log_2 2 = \cdots \end{equation*}

    Or, with your method: \log_2 1 - \log_2 8 = \log_2 1 - \log_2 2^3 = \cdots

    Note that \log_{\text{anything}} 1 = 0
    Yeah its been a while since i've done logs, so i've pretty much forgot all of the rules. For this question, could you just simply type log2(1/8) into a calculator and put whatever comes up as the answer?
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    (Original post by jessyjellytot14)
    Yeah its been a while since i've done logs, so i've pretty much forgot all of the rules. For this question, could you just simply type log2(1/8) into a calculator and put whatever comes up as the answer?
    Well, you could - but that would be cheating and you'd learn absolutely nothing. Why don't you google up the logarithm rules and learn them before attempting the homework? You can then check your answers to such questions by doing it manually, then by checking it with a calculator.
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    (Original post by Zacken)
    Well, you could - but that would be cheating and you'd learn absolutely nothing. Why don't you google up the logarithm rules and learn them before attempting the homework? You can then check your answers to such questions by doing it manually, then by checking it with a calculator.
    Yep that will probably be a good idea. Btw how would you sketch a graph of y= logax ?
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    (Original post by jessyjellytot14)
    Yep that will probably be a good idea. Btw how would you sketch a graph of y= logax ?
    I know that there will be a root at x=1, there's a vertical asymptote at x=0 where the function tends to negative infinity. I can tell that the function increases very rapidly in the interval from 0 to 1 and then slows down and increases rather slowly (i.e: decreasing gradient) after x=1, so that'll let me piece together a sketch of it.
 
 
 
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