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Question Number 96242 by joki last updated on 30/May/20

$$\mathrm{find}\mathrm{x}\mathrm{if}{4}^{\mathrm{x}}+6^{\mathrm{x}}=9^{\mathrm{x}}\mathrm{findx}?$$

Answered by john santu last updated on 30/May/20

$${\left(\frac{4}{9}\right)}^x+{\left(\frac{6}{9}\right)}^x=1$$$${\left(\frac{2}{3}\right)}^{2x}+{\left(\frac{2}{3}\right)}^x=1$$$$t^2+t-1=0,\mathrm{t}={\left(\frac{2}{3}\right)}^x$$$$t=\frac{-1+\sqrt{5}}{2}\Rightarrow x=\log {}_{\left(\frac{2}{3}\right)}\left(\frac{\sqrt{5}-1}{2}\right)$$

Answered by 1549442205 last updated on 31/May/20

$$\mathrm{Dividing}\mathrm{two}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{by}$$$$6^{\mathrm{x}}\mathrm{we}\mathrm{get}{\left(\frac{2}{3}\right)}^{\mathrm{x}}+1={\left(\frac{3}{2}\right)}^{\mathrm{x}}.\mathrm{Putting}{\left(\frac{2}{3}\right)}^{\mathrm{x}}$$$$=\mathrm{t}(\mathrm{t}>0),\mathrm{we}\mathrm{obtain}:\mathrm{t}-\frac{1}{\mathrm{t}}+1=0\iff {\mathrm{t}}^2+\mathrm{t}-1=0$$$$\mathrm{t}=\frac{-1+\sqrt{5}}{2}.\mathrm{From}\mathrm{that}{\left(\frac{2}{3}\right)}^{\mathrm{x}}=\frac{\sqrt{5}-1}{2}$$$$\mathrm{xln}\left(\frac{2}{3}\right)=\ln \frac{\sqrt{5}-1}{2}\Rightarrow \mathrm{x}=\frac{\ln \frac{\sqrt{5}-1}{2}}{\ln \frac{2}{3}}\approx 1.1868$$$$$$

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