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Question Number 5685 by sanusihammed last updated on 24/May/16

$$Findthevalueofx.$$$$$$$$9^x=6^x+4^x$$$$$$$$Pleasehelp.$$

Answered by prakash jain last updated on 24/May/16

$$\mathrm{Divide}\mathrm{by}{4}^x$$$${\left(\frac{9}{4}\right)}^x={\left(\frac{6}{4}\right)}^x+1$$$${\left(\frac{3}{2}\right)}^{2x}-{\left(\frac{3}{2}\right)}^x-1=0$$$${\left(\frac{3}{2}\right)}^x=u$$$$u^2-u-1=0$$$$\Rightarrow u=\frac{1\pm \sqrt{5}}{2}$$$$\mathrm{for}\mathrm{real}x,u>0\Rightarrow u=\frac{1+\sqrt{5}}{2}$$$${\left(\frac{3}{2}\right)}^x=\frac{1+\sqrt{5}}{2}$$$$x(\ln 3-\ln 2)=\ln (1+\sqrt{5})-\ln 2$$$$x=\frac{\ln (1+\sqrt{5})-\ln 2}{\ln 3-\ln 2}$$

Commented by sanusihammed last updated on 24/May/16

$$Thankyouforhourhelp.$$

Commented by Rasheed Soomro last updated on 24/May/16

$${\mathbb{N}}^{\mathbf{i}}\mathrm{CE}!$$

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