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Question Number 149192 by Samimsultani last updated on 03/Aug/21

Answered by prakash jain last updated on 03/Aug/21

((√3)+(√2))^x =u  ((√3)−(√2))^x =(1/u)  u−(1/u)=40(√6)  u^2 −40(√6)u−1=0  u=20(√6)±49  ((√3)+(√2))^x =20(√6)+49=25+20(√6)+2^2 6  =(5+2(√6))^2 =(3+2(√6)+2)^2 =((√3)+(√2))^4   x=4

$$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{{x}} ={u} \\ $$$$\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)^{{x}} =\frac{\mathrm{1}}{{u}} \\ $$$${u}−\frac{\mathrm{1}}{{u}}=\mathrm{40}\sqrt{\mathrm{6}} \\ $$$${u}^{\mathrm{2}} −\mathrm{40}\sqrt{\mathrm{6}}{u}−\mathrm{1}=\mathrm{0} \\ $$$${u}=\mathrm{20}\sqrt{\mathrm{6}}\pm\mathrm{49} \\ $$$$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{{x}} =\mathrm{20}\sqrt{\mathrm{6}}+\mathrm{49}=\mathrm{25}+\mathrm{20}\sqrt{\mathrm{6}}+\mathrm{2}^{\mathrm{2}} \mathrm{6} \\ $$$$=\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} =\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{2}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{4}} \\ $$$${x}=\mathrm{4} \\ $$

Commented by prakash jain last updated on 03/Aug/21

Yes. x=4 is only correct solution.  My mistake.  20(√6)−49 does not give any solution

$$\mathrm{Yes}.\:{x}=\mathrm{4}\:\mathrm{is}\:\mathrm{only}\:\mathrm{correct}\:\mathrm{solution}. \\ $$$$\mathrm{My}\:\mathrm{mistake}. \\ $$$$\mathrm{20}\sqrt{\mathrm{6}}−\mathrm{49}\:\mathrm{does}\:\mathrm{not}\:\mathrm{give}\:\mathrm{any}\:\mathrm{solution} \\ $$

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