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Question Number 129451 by bemath last updated on 15/Jan/21

$$\log {}_{12}(\sqrt{\mathrm{x}}+\sqrt[4]{\mathrm{x}})=\log {}_9\left(\sqrt{\mathrm{x}}\right)$$$$\mathrm{x}=?$$

Answered by liberty last updated on 16/Jan/21

$$\log {}_{12}(\sqrt{\mathrm{x}}+\sqrt[4]{\mathrm{x}})=\log {}_3\left(\sqrt[4]{\mathrm{x}}\right)$$$$\mathrm{let}\mathrm{x}={\mathrm{t}}^4\Rightarrow \frac{\ln ({\mathrm{t}}^2+\mathrm{t})}{\ln 12}=\frac{\ln \left(\mathrm{t}\right)}{\ln 3}$$$$\ln 3.\ln ({\mathrm{t}}^2+\mathrm{t})=\ln 12.\ln \left(\mathrm{t}\right)$$$$\ln 3.\ln ({\mathrm{t}}^2+\mathrm{t})=(\ln 3+\ln 4).\ln \left(\mathrm{t}\right)$$$$\mathrm{holds}\mathrm{for}\mathrm{t}=3;\mathrm{then}\mathrm{x}=3^4=81$$$$$$

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