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Thread: How should I solve these equations?

  1. #1
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    How should I solve these equations?

    Hey all,
    I am trying to solve these 2 equations but I can't.
    $\displaystyle 27^x + 12^x = 2 * 8^3$
    and,
    $\displaystyle 3 * 16 ^x + 37 * 36^x = 26 * 81^x$
    thanks in advance.
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  2. #2
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    Quote Originally Posted by mehdi View Post
    Hey all,
    I am trying to solve these 2 equations but I can't.
    $\displaystyle 27^x + 12^x = 2 * 8^3$
    and,
    $\displaystyle 3 * 16 ^x + 37 * 36^x = 26 * 81^x$
    thanks in advance.
    I can show you how to do the 2nd one - but it is kind of tricky:

    $\displaystyle 3 * 16 ^x + 37 * 36^x = 26 * 81^x$

    $\displaystyle 37 \cdot 6^{2x}+3 \cdot 2^{4x} = 26 \cdot 3^{4x}$

    $\displaystyle 37 \cdot 2^{2x} \cdot 3^{2x}+3 \cdot 2^{4x} = 26 \cdot 3^{4x}$ divide by $\displaystyle 2^{2x}$

    $\displaystyle 37 \cdot 3^{2x}+3 \cdot 2^{2x} = 26 \cdot 3^{4x} \cdot \dfrac1{2^{2x}}$ divide by $\displaystyle 3^{2x}$

    $\displaystyle 37 +3 \cdot \left(\dfrac23\right)^{2x} = 26 \cdot 3^{2x} \cdot \dfrac1{2^{2x}} = 26 \cdot \left(\dfrac32\right)^{2x}$

    Now substitute $\displaystyle y = \left(\dfrac32\right)^{2x}$. The equation becomes:

    $\displaystyle 37+3 \cdot \dfrac1y=26 \cdot y$ multiply by y:

    $\displaystyle 26y^2-37y-3=0$

    yields $\displaystyle y = \frac32~\vee~y=-\frac1{13}$

    Since y is a power with a positive base the 2nd solution can't be correct.

    Now re-substitute:

    $\displaystyle \left(\dfrac32\right)^{2x} = \dfrac32$

    That means $\displaystyle 2x = 1~\implies~\boxed{x = \dfrac12}$
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  4. #4
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    Hello, mehdi!

    Another approach to the second problem . . .


    $\displaystyle 3\cdot16^x + 37\cdot36^x \:=\: 26\cdot81^x$

    We have: .$\displaystyle 3(2^4)^x + 37(2^2\!\cdot\!3^3)^x - 26(3^4)^x \;=\;0$

    . . . . . . .$\displaystyle 3(2^{4x}) + 37(2^{2x})(3^{2x}) - 26(3^{4x}) \;=\;0 $

    Factor: . $\displaystyle \bigg[2^{2x} + 13\!\cdot\!3^{2x}\bigg]\,\bigg[3\!\cdot\!2^{2x} - 2\!\cdot\!3^{2x}\bigg] \;=\;0 $



    We have two equations to solve:


    $\displaystyle 2^{2x} + 13\!\cdot\!3^{2x} \:=\:0 \quad\Rightarrow\quad 2^{2x} \:=\:\text{-}13\!\cdot\!3^{2x}$

    . . $\displaystyle \dfrac{2^{2x}}{3^{2x}} \:=\:\text{-}13 \quad\Rightarrow\quad \left(\dfrac{2}{3}\right)^{2x} \:=\:\text{-}13\;\hdots\;\text{ no real roots}$


    $\displaystyle 3\!\cdot\!2^{2x} - 2\!\cdot\!3^{2x} \:=\:0 \quad\Rightarrow\quad 3\!\cdot\!2^{2x} \:=\:2\!\cdot\!3^{2x}$

    . . $\displaystyle \dfrac{2^{2x}}{3^{2x}} \:=\:\dfrac{2}{3} \quad\Rightarrow\quad \left(\dfrac{2}{3}\right)^{2x} =\:\left(\dfrac{2}{3}\right)^1 \quad\Rightarrow\quad 2x \:=\:1$

    . . $\displaystyle \text{Therefore: }\:\boxed{x \:=\:\frac{1}{2}}$

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