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Question Number 96636 by O Predador last updated on 03/Jun/20

Commented by hknkrc46 last updated on 03/Jun/20

$$5^x-{(9,8)}^x=7^x\Rightarrow 5^x-{\left(\frac{49}{5}\right)}^x=7^x$$$$\Rightarrow 5^x-\frac{7^{2x}}{5^x}=7^x\Rightarrow 5^{2x}-7^{2x}=5^x\cdot 7^x$$$$\Rightarrow \frac{5^{2x}-7^{2x}}{5^x\cdot 7^x}=1\Rightarrow \frac{5^x}{7^x}-\frac{7^x}{5^x}=1$$$$\frac{5^x}{7^x}={\left(\frac{5}{7}\right)}^x=t\Rightarrow t-\frac{1}{t}=1\Rightarrow t^2-t-1=0$$$$\Rightarrow t=\frac{-(-1)\mp \sqrt{{(-1)}^2-4\cdot 1\cdot (-1)}}{2\cdot 1}$$$$\Rightarrow {\left(\frac{5}{7}\right)}^x=\frac{1\mp \sqrt{5}}{2}\Rightarrow 0<{\left(\frac{5}{7}\right)}^x=\frac{1+\sqrt{5}}{2}$$$$\Rightarrow \ln {\left(\frac{5}{7}\right)}^x=\ln \frac{1+\sqrt{5}}{2}$$$$\Rightarrow x\ln \left(\frac{5}{7}\right)=\ln \frac{1+\sqrt{5}}{2}\Rightarrow x=\frac{\ln \frac{1+\sqrt{5}}{2}}{\ln \frac{5}{7}}$$

Answered by mr W last updated on 03/Jun/20

$${\left(\frac{1}{1.4}\right)}^x-{(1.4)}^x=1$$$${(1.4)}^{2x}+{(1.4)}^x-1=0$$$$1.4^x=\frac{\sqrt{5}-1}{2}$$$$\Rightarrow x=\frac{\ln (\sqrt{5}-1)-\ln 2}{\ln 7-\ln 5}\approx -1.4302$$

Commented by O Predador last updated on 03/Jun/20

$$\mathbf{or}{\mathbf{log}}_{\frac{7}{5}}\left(\frac{-1+\sqrt{5}}{2}\right)?$$

Commented by mr W last updated on 03/Jun/20

$$yes$$

Commented by O Predador last updated on 03/Jun/20

$$\mathbf{Thank}\mathbf{you}!$$

Answered by 1549442205 last updated on 03/Jun/20

$$\mathrm{Equation}\mathrm{is}\mathrm{equivalent}\mathrm{to}$$$$5^{\mathrm{x}}-{\left(\frac{49}{5}\right)}^{\mathrm{x}}=7^{\mathrm{x}}\iff 5^{2\mathrm{x}}-7^{2\mathrm{x}}=7^{\mathrm{x}}\times 5^{\mathrm{x}}\left(1\right)$$$$\iff {\left(\frac{5}{7}\right)}^{\mathrm{x}}-{\left(\frac{7}{5}\right)}^{\mathrm{x}}=1$$$$\mathrm{et}{\left(\frac{7}{5}\right)}^{\mathrm{x}}=\mathrm{y}(\mathrm{y}>0)$$$$\Rightarrow \frac{1}{\mathrm{y}}-\mathrm{y}=1\iff {\mathrm{y}}^2+\mathrm{y}-1=0$$$$\Rightarrow \mathrm{y}=\frac{-1+\sqrt{5}}{2}\Rightarrow {\left(\frac{7}{5}\right)}^{\mathrm{t}}=\frac{-1+\sqrt{5}}{2}$$$$\Rightarrow \mathrm{x}=\frac{\ln \left(\frac{-1+\sqrt{5}}{2}\right)}{\ln \frac{7}{5}}\approx -1.43016$$

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