第03期 - 底数与指数相同的方程

$\text{解方程：} n^9 = 3^n$

$\text{两边取} \frac{1}{n} \text{次方：} (n^9)^{\frac{1}{n}} = (3^n)^{\frac{1}{n}}$

$n^{\frac{9}{n}} = 3$

$\text{两边取} 9 \text{次方：} (n^{\frac{1}{n}})^9 = 3$

$n^{\frac{1}{n}} = 3^{\frac{1}{9}}$

$\text{写成指数形式：} n^{\frac{1}{n}} = (3^3)^{\frac{1}{27}}$

$n^{\frac{1}{n}} = 27^{\frac{1}{27}}$

$\text{由此得出：} n = 27$