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

2020-5-2 258

Solve

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

• 2020-5-3
2

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

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

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