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楼主 |
发表于 2020-12-3 16:36
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题: 解方程 \(\;4^x+6^x=9^x\;(x\in\mathbb{R})\)
解: 两边除以\(\,4^x\,\)得\(\,\small 1+\big(\dfrac{3}{2}\big)^x=\big(\dfrac{9}{4}\big)^x,\;\;y^2-y-1=0,\;(y=\small\big(\dfrac{3}{2}\big)^x)\)
\({\small\big(\dfrac{3}{2}\big)^x=\dfrac{1+\sqrt{5}}{2}},\;\;x=\dfrac{\ln\frac{1+\sqrt{5}}{2}}{\ln\frac{3}{2}}.\) |
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