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

Question Number 131890 by Study last updated on 09/Feb/21

$${log}_{\mathrm{2}} {x}+{log}_{\mathrm{3}} {x}=\mathrm{1}\:\:\:\:\:{x}=? \\ $$

Answered by Raxreedoroid last updated on 09/Feb/21

$$\frac{{log}_{\mathrm{3}} {x}}{{log}_{\mathrm{3}} \mathrm{2}}+{log}_{\mathrm{3}} {x}=\mathrm{1} \\ $$$${log}_{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }=\mathrm{1} \\ $$$${x}=\mathrm{3} \\ $$

Commented by liberty last updated on 09/Feb/21

$$\mathrm{wrong} \\ $$

Commented by Raxreedoroid last updated on 09/Feb/21

$$\mathrm{ah}\:\mathrm{sorry}\:\mathrm{a}\:\mathrm{mistake} \\ $$$$\frac{{log}_{\mathrm{3}} {x}}{{log}_{\mathrm{3}} \mathrm{2}}+{log}_{\mathrm{3}} {x}=\mathrm{1} \\ $$$${log}_{\mathrm{3}} {x}\:\centerdot\left(\mathrm{1}+{log}_{\mathrm{3}} \mathrm{2}\right)=\mathrm{1} \\ $$$${log}_{\mathrm{3}} {x}=\frac{\mathrm{1}}{\mathrm{1}+{log}_{\mathrm{3}} \mathrm{2}} \\ $$$${x}=\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{1}+{log}_{\mathrm{3}} \mathrm{2}}} \approx\mathrm{1}.\mathrm{9613}\:… \\ $$

Commented by EDWIN88 last updated on 09/Feb/21

$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\Rightarrow\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\:\left[\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)}\:\right]=\mathrm{1} \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\:\left[\:\frac{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{1}}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)}\:\right]=\mathrm{1} \\ $$$$\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{6}\right)}\:\Rightarrow\mathrm{x}\:=\:\mathrm{3}^{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)} \approx\mathrm{1}.\mathrm{529592} \\ $$$$\mathrm{3}^{\mathrm{log}\:_{\mathrm{6}} \mathrm{2}} \\ $$$$\mathrm{1}.\mathrm{529592} \\ $$

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