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Math Help - Help with gr 12 math?

  1. #1
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    Help with gr 12 math?

    How would I do this question below:

    11. Solve and check for extraneous roots. Where necessary, round answers to 2 decimal places.

    3^2x = 7(3^x)-12

    The 2x is the power of 3.

    I have no idea how to even start that. I tried dividing it by 7 first. Didn't work. Then I tried bringing over the 12 and that didn't work either.
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  2. #2
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     3^{2x} = 7(3^x)-12

     (3^x)^2 = 7(3^x)-12

    make 3^x = a

     a^2 = 7a-12

     a^2 - 7a+12=0

     (a-3)(a-4)=0

     a=3,4

     3^x=3,4

     x=1,\log_34
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  3. #3
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    Quote Originally Posted by pickslides View Post
     3^{2x} = 7(3^x)-12

     (3^x)^2 = 7(3^x)-12

    make 3^x = a

     a^2 = 7a-12

     a^2 - 7a+12=0

     (a-3)(a-4)=0

     a=3,4

     3^x=3,4

     x=1,\log_34
    thank you. but how did you get 3^{2x} to letting a be a^2 when it was 2x and not x^2?
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  4. #4
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     3^{2x} = (3^x)^2

    comes from the index law

     (a^m)^n = a^{m\times n}
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    Quote Originally Posted by pickslides View Post
     3^{2x} = (3^x)^2

    comes from the index law

     (a^m)^n = a^{m\times n}
    thank you. So when I applied when I learned from you to my next question I got it wrong. This is what I did.

    Solve and check for extraneous roots.

    5^x = 3 - 5^{2x}

    Let a be 5^x

    a = 3 - a^2
     a^2+a-3
     a = -2.23 or 1.232

    The answer in the book is a = 0.16
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  6. #6
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    Quote Originally Posted by Barthayn View Post

    a = 3 - a^2
     a^2+a-3
     a = -2.23 or 1.232
    assuming your solutions for a are correct you must then re-visit what you made a

     5^x = -2.23 or 5^x= 1.232

    The first cannot be solved because you can't have the log of a negative number so you have only one solution x= \log_51.232
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  7. #7
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    according to the answer book the answer is 0.16. What did I do wrong?
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  8. #8
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    Quote Originally Posted by Barthayn View Post
    according to the answer book the answer is 0.16. What did I do wrong?
    Using the quadratic equation, I found roots to be:

    1.303
    -2.303

    So, when solving  a^2 + a -3 , something went awry.

    What pickslides said about negative logs holds true here, so -2.303 isn't a solution.

    When you solve for x

    <br />
5^x = 1.303
    using logs, x = .1644

    which is probably what the book is talking about.
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  9. #9
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    Maybe this is a longwinded way to do it:

    5^x=3-5^{2x}

    let 5^x=a

    a^2+a-3=0

    \left (a+\frac{1+\sqrt{13}}{2}\right )\left (a+\frac{1-\sqrt{13}}{2}\right )=0

    5^x=\frac{\sqrt{13}-1}{2} (as previously stated by Pickslides, you can disregard the other solution, since you can't take the log of a negative value)

    x=\frac{log(\frac{1}{2}(\sqrt{13}-1))}{log5}

    x\approx 0.16
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  10. #10
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    Quote Originally Posted by pickslides View Post
    assuming your solutions for a are correct you must then re-visit what you made a

     5^x = -2.23 or 5^x= 1.232

    The first cannot be solved because you can't have the log of a negative number so you have only one solution x= \log_51.232
    Quote Originally Posted by Stroodle View Post
    Maybe this is a longwinded way to do it:

    5^x=3-5^{2x}

    let 5^x=a

    a^2+a-3=0

    \left (a+\frac{1+\sqrt{13}}{2}\right )\left (a+\frac{1-\sqrt{13}}{2}\right )=0

    5^x=\frac{\sqrt{13}-1}{2} (as previously stated by Pickslides, you can disregard the other solution, since you can't take the log of a negative value)

    x=\frac{log(\frac{1}{2}(\sqrt{13}-1))}{log5}

    x\approx 0.16
    thank you all. First I did not add the 1 to the 12 as well then log know variable to get x by its self.
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