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Question Number 55969 by Muhammad bello last updated on 07/Mar/19

$$\mathbf{solve}\mathbf{for}\mathbf{x}\mathrm{\&}\mathbf{y}$$ $$$$ $${\mathbf{x}}^{\mathbf{logy}}=4$$ $$$$ $$\mathbf{xy}=40$$

Answered by MJS last updated on 07/Mar/19

$$x^{\ln y}=4$$ $$\ln x\ln y=\ln 4\left(1\right)$$ $$xy=40$$ $$\ln x+\ln y=\ln 40\left(2\right)$$ $$\left(1\right)pq=a\Rightarrow q=\frac{a}{p}$$ $$\left(2\right)p+q=b\Rightarrow p+\frac{a}{p}=b\Rightarrow p^2-bp+a=0\Rightarrow$$ $$\Rightarrow p=\frac{b}{2}\pm \frac{\sqrt{b^2-4a}}{2}$$ $$\Rightarrow q=\frac{b}{2}\mp \frac{\sqrt{b^2-4a}}{2}$$ $$\ln x=\frac{\ln 40}{2}\pm \frac{\sqrt{{\left(\ln 40\right)}^2-4\ln 4}}{2}$$ $$\ln y=\frac{\ln 40}{2}\mp \frac{\sqrt{{\left(\ln 40\right)}^2-4\ln 4}}{2}$$ $$\Rightarrow (x\approx 1.52913\wedge y\approx 26.1587)\vee (x\approx 26.1587\wedge y\approx 1.52913)$$

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