Question Number 5685 by sanusihammed last updated on 24/May/16
$${Find}\:{the}\:{value}\:{of}\:{x}\:. \\ $$$$ \\ $$$$\mathrm{9}^{{x}} \:=\:\mathrm{6}^{{x}} \:+\:\mathrm{4}^{{x}} \\ $$$$ \\ $$$${Please}\:{help}. \\ $$
Answered by prakash jain last updated on 24/May/16
$$\mathrm{Divide}\:\mathrm{by}\:\mathrm{4}^{{x}} \\ $$$$\left(\frac{\mathrm{9}}{\mathrm{4}}\right)^{{x}} =\left(\frac{\mathrm{6}}{\mathrm{4}}\right)^{{x}} +\mathrm{1} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}{x}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} −\mathrm{1}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} ={u} \\ $$$${u}^{\mathrm{2}} −{u}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{u}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{real}\:{x},{u}>\mathrm{0}\Rightarrow{u}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}\left(\mathrm{ln}\:\mathrm{3}−\mathrm{ln}\:\mathrm{2}\right)=\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)−\mathrm{ln}\:\mathrm{2} \\ $$$${x}=\frac{\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)−\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}−\mathrm{ln}\:\mathrm{2}} \\ $$
Commented by sanusihammed last updated on 24/May/16
$${Thank}\:{you}\:{for}\:{hour}\:{help}. \\ $$
Commented by Rasheed Soomro last updated on 24/May/16
$$\mathbb{N}^{\boldsymbol{\mathrm{i}}} \mathrm{CE}! \\ $$