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Thread: Solve for x: 3^(2x) - 5(3^x) = -6

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    Question Solve for x: 3^(2x) - 5(3^x) = -6

    Solve for x: 3^(2x) - 5(3^x) = -6

    Asap please
    some explanation would be appreciated
    thank you


    Edit: the correct answers are 1 and 0.631, but why and how?
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    Quote Originally Posted by incognito_301 View Post
    Solve for x: 3^(2x) - 5(3^x) = -6

    Asap please
    some explanation would be appreciated
    thank you


    Edit: the correct answers are 1 and 0.631, but why and how?
    Here's a hint: note that $\displaystyle 3^{2x} = (3^x)^2$

    now let $\displaystyle y = 3^x$

    then we have $\displaystyle y^2 - 5y = -6$

    that is, $\displaystyle y^2 - 5y + 6 = 0$

    this is just a quadratic equation. you can solve for $\displaystyle y$ and then replace it with $\displaystyle 3^x$ and then solve for $\displaystyle x$
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    Quote Originally Posted by incognito_301 View Post
    Solve for x: 3^(2x) - 5(3^x) = -6

    Asap please
    some explanation would be appreciated
    thank you


    Edit: the correct answers are 1 and 0.631, but why and how?
    $\displaystyle 3^{2x} - 5(3^x) - 6 = 0 $

    $\displaystyle (3^{x})^2 - 5(3^x) - 6 = 0 $

    It's a quadratic. Solve it.
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  4. #4
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    Quote Originally Posted by incognito_301 View Post
    Solve for x: 3^(2x) - 5(3^x) = -6

    Asap please
    [snip]
    I would have thought you would want someone more competent than a sap to help you.

    Fortunately you got that competent help. But I wouldn't make a habit of asking for a sap to help you .....
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    LOL a sap
    wow amazing forum! didnt believe you replied so fast. thanks.

    alright, so





    $\displaystyle A^2-5(A)-6=0$

    A = -1 and A = 6

    $\displaystyle 3^x = -1$
    $\displaystyle 3^x = 6$

    then l$\displaystyle og 6/log3 = x$
    and $\displaystyle log -1/log3 = x$

    i dont get the correct answers (1 and 0.631).
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by incognito_301 View Post
    LOL a sap
    wow amazing forum! didnt believe you replied so fast. thanks.

    alright, so





    $\displaystyle A^2-5(A)-6=0$

    A = -1 and A = 6

    $\displaystyle 3^x = -1$
    $\displaystyle 3^x = 6$

    then l$\displaystyle og 6/log3 = x$
    and $\displaystyle log -1/log3 = x$

    i dont get the correct answers (1 and 0.631).
    as i said, it is a +6 not -6. when you move -6 over the equal sign, it becomes positive. since you are adding 6 to both sides

    but other than that, your method is correct

    (by the way, log(-1) is not defined in the real numbers. you'd put "no solution" in that case)
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