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  1. #1
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    Logarithm question

    Hi there, I'm having difficulty with this question:

    Two variables, x and y, are connected by the law y = a^x. The graph of log4y against x is a straight line passing through the origin and the point A(6,3). Find the value of a.

    Thanks in advance
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  2. #2
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    Quote Originally Posted by HighlyFlammable View Post
    Hi there, I'm having difficulty with this question:

    Two variables, x and y, are connected by the law y = a^x. The graph of log4y against x is a straight line passing through the origin and the point A(6,3). Find the value of a.

    Thanks in advance
    If we make a table of values you will see that

    \begin{array}{l|c|r} x & y & \log_{4}(y) \\  \hline 0 & 1 & 0 \\ 1 & a & log_{4}(a ) \\  2 & a^2 & 2log_{4}(a ) \\ 3 & a^3 & 3log_{4}(a ) \\ 4 & a^4 & 4log_{4}(a ) \\ 5 & a^5 & 5log_{4}(a ) \\ 6 & a^6 & 6log_{4}(a ) \\ \end{array}

    Now we can pick the correct ordered pair to get

    6log_{4}(a )=3 \iff log_{4}(a ) =\frac{1}{2} \iff a=4^{\frac{1}{2}}=2
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  3. #3
    Super Member Quacky's Avatar
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    Quote Originally Posted by HighlyFlammable View Post
    Hi there, I'm having difficulty with this question:

    Two variables, x and y, are connected by the law y = a^x. The graph of log4y against x is a straight line passing through the origin and the point A(6,3). Find the value of a.

    Thanks in advance
    The wording of the question confused me too.

    My initial thoughts were this:

    Log_4(y)=x
    y=4^x
    a^x=4^x
    a=4

    But this does not satisfy either of the given points.Therefore it must be wrong.

    The reason is that the x's and y's in the question seemed slightly inconsistent with their meaning. I guess I need to brush up on my understanding of language!

    I decided to start with the straight line passing through (6,3) and the origin. The gradient of this line is \frac{3}{6}=\frac{1}{2} The y-intercept is 0.

    So the line they are referring to is y=\frac{1}{2}x.

    Now, I assumed then that:

    Log_4{(y)}=\frac{1}{2}x

    ...which, when you solve for a, gives the same answer as TES.

    After rereading the question, I realised that it does make sense but that I was incapable of grasping its meaning.
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