[Problem of the day][数学挑战][Powers and Indices][指数和幂运算][2] Mathematics Algebra

casperyc 2020-5-2 258

Solve

$$ \huge 4^x + 6^x = 9^x $$

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    $$ \huge 4^x + 6^x = 9^x $$

    Key is first to identify that for $4,6$ and $9$, we only have two factors, $2$ and $3$. Hence if we divide both sides by $4^x$, we get

    $$ \huge 1 + \left(\frac{3}{2} \right)^x = \left(\frac{3}{2} \right)^{2x} $$

    We end up with a quadratic equation, solving for $\left(\frac{3}{2} \right)^x$ yields

    $$ \huge \left(\frac{3}{2} \right)^x = \frac{1+\sqrt{5}}{2} $$

    It follows that

    $$ \huge x = \frac{\ln \left( \frac{1+\sqrt{5}}{2} \right)  }{\ln \left( \frac32 \right)} $$

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