Question Number 164806 by saboorhalimi last updated on 22/Jan/22
Commented by cortano1 last updated on 22/Jan/22
$$\:\mathrm{log}\:_{\mathrm{12}} \left(\mathrm{3}\right)={a}\:\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{12}\right)=\frac{\mathrm{1}}{{a}} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{2}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)=\frac{\mathrm{1}}{{a}}\: \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)=\:\frac{\mathrm{1}−{a}}{\mathrm{2}{a}}\: \\ $$$$\Rightarrow\frac{\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{2}\right)}{\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right)}\:=\:\frac{\mathrm{1}−{a}}{\mathrm{2}{a}} \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{2}\right)=\left(\frac{\mathrm{1}−{a}}{\mathrm{2}{a}}\right)\:\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right) \\ $$$$ \\ $$
Commented by cortano1 last updated on 22/Jan/22
$$\:\mathrm{log}\:_{\mathrm{12}} \left(\mathrm{3}\right)=\frac{\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right)}{\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{12}\right)}\:=\:{a} \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right)={a}\left(\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right)+\mathrm{2log}\:_{\mathrm{5}} \left(\mathrm{2}\right)\right) \\ $$$$\Rightarrow\left(\mathrm{1}−{a}\right)\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right)=\mathrm{2}{a}\:\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{2}\right)=\frac{\left(\mathrm{1}−{a}\right)}{\mathrm{2}{a}}\:\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right) \\ $$
Commented by mr W last updated on 22/Jan/22
$${is}\:\mathrm{log}_{\mathrm{5}} \:\mathrm{3}\:{better}\:{than}\:\mathrm{log}_{\mathrm{5}} \:\mathrm{2}\:? \\ $$
Commented by cortano1 last updated on 22/Jan/22
$${yes} \\ $$
Commented by mr W last updated on 22/Jan/22
$${and}\:{how}\:{to}\:{get}\:\mathrm{log}_{\mathrm{5}} \:\mathrm{3}? \\ $$
Commented by cortano1 last updated on 22/Jan/22
$${by}\:{calculator}\: \\ $$
Commented by cortano1 last updated on 22/Jan/22
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Answered by MJS_new last updated on 22/Jan/22
$$\mathrm{log}_{\mathrm{12}} \:\mathrm{3}\:={a}\:\Leftrightarrow\:\mathrm{log}_{\mathrm{5}} \:\mathrm{2}\:={b} \\ $$$$\mathrm{or}\:\mathrm{else}\:\mathrm{solve}\:\mathrm{this}:\:\sqrt[{\mathrm{17}}]{\mathrm{29}}=\alpha,\:\sqrt[{\mathrm{13}}]{\mathrm{7}}=? \\ $$