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Math Help - log equations

  1. #1
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    log equations

    I just did a past paper, and heres the two questions i couldn't do.



    For the part 'i' question, i think its something to do with the unit circle, but im not sure. This is a non-calculator paper so i have to find it some other way. Or it might just be a question of learning it off by heart?

    For the part 'ii' question, i can't seem to get it to work. I think i know the rules of logs pretty well. I started by dividing both terms by eachother as there is a - sign between them. Could someone give me help on that.

    Thanks in advanced.
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  2. #2
    Moo
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    Quote Originally Posted by nugiboy View Post
    I just did a past paper, and heres the two questions i couldn't do.



    For the part 'i' question, i think its something to do with the unit circle, but im not sure. This is a non-calculator paper so i have to find it some other way. Or it might just be a question of learning it off by heart?
    Because \tan 300=\tan (360-60)
    Two possibilities :
    - you transform tan into sin/cos :
    you use the unit circle
    you use the formulaes cos(a-b) and sin(a-b)
    - you know the formula tan(a-b)


    For the part 'ii' question, i can't seem to get it to work. I think i know the rules of logs pretty well. I started by dividing both terms by eachother as there is a - sign between them. Could someone give me help on that.

    Thanks in advanced.
    We'll take care of the left hand side of the equation.
    First of all, put every constant inside the logarithm.

    \begin{aligned} 2 \log_a \frac{x^3}{4} &=\log_a \left[\left(\frac{x^3}{4}\right)^2\right] \\<br />
&=\log_a \frac{x^6}{2^4} \end{aligned}

    Now you can divide :

    \begin{aligned} \log_a x^{10}-\log_a \frac{x^6}{2^4} &=\log_a \left(x^{10} \cdot \frac{2^4}{x^6} \right) \\<br />
&=\log_a (2^4 \cdot x^4) \\<br />
&=\dots \end{aligned}

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  3. #3
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    Hello, nugiboy!

    2) Show that: . \log_a\!\left(x^{10}\right) - 2\log_a\!\left(\frac{x^3}{4}\right) \;=\;4\log_a(2x)

    The left side is: . \log_a\!\left(x^{10}\right) - \log_a\!\left(\frac{x^3}{4}\right)^2 \;\;=\; \;\log_a\!\left(x^{10}\right) - \log_a\!\left(\frac{x^6}{16}\right)

    . . = \;\;\log_a\!\left(\frac{x^{10}}{\frac{x^6}{16}}\ri  ght) \;\;=\;\;\log_a\!\left(16x^4\right) \;\;=\;\;\log_a(2x)^4 \;\;=\;\;4\log_a(2x)

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, nugiboy!


    The left side is: . \log_a\!\left(x^{10}\right) - \log_a\!\left(\frac{x^3}{4}\right)^2 \;\;=\; \;\log_a\!\left(x^{10}\right) - \log_a\!\left(\frac{x^6}{16}\right)

    . . = \;\;\log_a\!\left(\frac{x^{10}}{\frac{x^6}{16}}\ri  ght) \;\;=\;\;\log_a\!\left(16x^4\right) \;\;=\;\;\log_a(2x)^4 \;\;=\;\;4\log_a(2x)

    Cheers man

    Thats exactly what i did, but got stuck near the end. I got up to

     <br />
\;\;\log_a\!\left(16x^4\right) \;\; <br />

    , but im not sure how that gets to the next bit.
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  5. #5
    Moo
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    Quote Originally Posted by nugiboy View Post
    Cheers man

    Thats exactly what i did, but got stuck near the end. I got up to

     <br />
\;\;\log_a\!\left(16x^4\right) \;\; <br />

    , but im not sure how that gets to the next bit.
    16=2^4

    -> 16x^4=(2x)^4

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