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Question-162814




Question Number 162814 by amin96 last updated on 01/Jan/22
Answered by MathsFan last updated on 01/Jan/22
 let  log_3 x=u → x=3^u    2^u =(√3^u )+1   2^u −(√3^u )=4−3   2^u =4  →  u=2   x=3^2     x=9
$$\:{let}\:\:{log}_{\mathrm{3}} {x}={u}\:\rightarrow\:{x}=\mathrm{3}^{{u}} \\ $$$$\:\mathrm{2}^{{u}} =\sqrt{\mathrm{3}^{{u}} }+\mathrm{1} \\ $$$$\:\mathrm{2}^{{u}} −\sqrt{\mathrm{3}^{{u}} }=\mathrm{4}−\mathrm{3} \\ $$$$\:\mathrm{2}^{{u}} =\mathrm{4}\:\:\rightarrow\:\:{u}=\mathrm{2} \\ $$$$\:{x}=\mathrm{3}^{\mathrm{2}} \: \\ $$$$\:{x}=\mathrm{9} \\ $$$$\: \\ $$
Answered by Exauce last updated on 01/Jan/22
              X =^(log_3 2) (√((√X) + 1))
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{X}}\:=\:^{\boldsymbol{{log}}_{\mathrm{3}} \mathrm{2}} \sqrt{\sqrt{\boldsymbol{{X}}}\:+\:\mathrm{1}} \\ $$

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