log-2-x-log-3-x-1-x-

Question Number 91862 by jagoll last updated on 03/May/20

$$\mathrm{log}_{\mathrm{2}} \left({x}\right)+\mathrm{log}_{\mathrm{3}} \left({x}\right)\:=\:\mathrm{1} \\ $$$${x}\:=? \\ $$

Commented by Tony Lin last updated on 03/May/20

((lnx)/(ln2))+((lnx)/(ln3))=1 lnx(((ln6)/(ln2×ln3)))=1 lnx=((ln2×ln3)/(ln6)) x=e^((ln2×ln3)/(ln6)) ≈1.53

$$\frac{{lnx}}{{ln}\mathrm{2}}+\frac{{lnx}}{{ln}\mathrm{3}}=\mathrm{1} \\ $$$${lnx}\left(\frac{{ln}\mathrm{6}}{{ln}\mathrm{2}×{ln}\mathrm{3}}\right)=\mathrm{1} \\ $$$${lnx}=\frac{{ln}\mathrm{2}×{ln}\mathrm{3}}{{ln}\mathrm{6}} \\ $$$${x}={e}^{\frac{{ln}\mathrm{2}×{ln}\mathrm{3}}{{ln}\mathrm{6}}} \approx\mathrm{1}.\mathrm{53} \\ $$

Commented by jagoll last updated on 03/May/20

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Commented by hmamarques1994@gmail.com last updated on 03/May/20