Question-169916

Question Number 169916 by mathlove last updated on 12/May/22

Answered by mahdipoor last updated on 12/May/22

ln(2^x^2 ×3^x )=ln(2^x^2 )+ln(3^x )=ln(6) ⇒ (ln2)x^2 +(ln3)x−ln6=0 ⇒ x=((−ln3±(√(ln^2 3+4.ln2.ln6)))/(2ln2)) , ln^2 3+4.ln2.ln6=ln^2 3+4.ln2.(ln3+ln2)= ln^2 3+4ln^2 2+4.ln2.ln3=(2ln2+ln3)^2 , x=((−ln3±(2ln2+ln3))/(2ln2))= { (1),((−1−((ln3)/(ln2))=−log_2 6)) :}

$${ln}\left(\mathrm{2}^{{x}^{\mathrm{2}} } ×\mathrm{3}^{{x}} \right)={ln}\left(\mathrm{2}^{{x}^{\mathrm{2}} } \right)+{ln}\left(\mathrm{3}^{{x}} \right)={ln}\left(\mathrm{6}\right)\:\Rightarrow \\ $$$$\left({ln}\mathrm{2}\right){x}^{\mathrm{2}} +\left({ln}\mathrm{3}\right){x}−{ln}\mathrm{6}=\mathrm{0}\:\Rightarrow \\ $$$${x}=\frac{−{ln}\mathrm{3}\pm\sqrt{{ln}^{\mathrm{2}} \mathrm{3}+\mathrm{4}.{ln}\mathrm{2}.{ln}\mathrm{6}}}{\mathrm{2}{ln}\mathrm{2}}\:, \\ $$$${ln}^{\mathrm{2}} \mathrm{3}+\mathrm{4}.{ln}\mathrm{2}.{ln}\mathrm{6}={ln}^{\mathrm{2}} \mathrm{3}+\mathrm{4}.{ln}\mathrm{2}.\left({ln}\mathrm{3}+{ln}\mathrm{2}\right)= \\ $$$${ln}^{\mathrm{2}} \mathrm{3}+\mathrm{4}{ln}^{\mathrm{2}} \mathrm{2}+\mathrm{4}.{ln}\mathrm{2}.{ln}\mathrm{3}=\left(\mathrm{2}{ln}\mathrm{2}+{ln}\mathrm{3}\right)^{\mathrm{2}} \\ $$$$, \\ $$$${x}=\frac{−{ln}\mathrm{3}\pm\left(\mathrm{2}{ln}\mathrm{2}+{ln}\mathrm{3}\right)}{\mathrm{2}{ln}\mathrm{2}}=\begin{cases}{\mathrm{1}}\\{−\mathrm{1}−\frac{{ln}\mathrm{3}}{{ln}\mathrm{2}}=−{log}_{\mathrm{2}} \mathrm{6}}\end{cases} \\ $$

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