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

	Answered by mahdipoor last updated on 12/May/22
	$ln(2^x^2 ×3^x )=ln(2^x^2 )+ln(3^x )=ln(6) ⇒ (ln2)x^2 +(ln3)x−ln6=0 ⇒ x=((−ln3±(√(ln^2 3+4.ln2.ln6)))/(2ln2)) , ln^2 3+4.ln2.ln6=ln^2 3+4.ln2.(ln3+ln2)= ln^2 3+4ln^2 2+4.ln2.ln3=(2ln2+ln3)^2 , x=((−ln3±(2ln2+ln3))/(2ln2))= { (1),((−1−((ln3)/(ln2))=−log_2 6)) :}$
	$l n (2^{x^{2}} \times 3^{x}) = l n (2^{x^{2}}) + l n (3^{x}) = l n (6) \Rightarrow$ $(l n 2) x^{2} + (l n 3) x - l n 6 = 0 \Rightarrow$ $x = \frac{- l n 3 \pm \sqrt{{l n}^{2} 3 + 4 . l n 2 . l n 6}}{2 l n 2},$ ${l n}^{2} 3 + 4 . l n 2 . l n 6 = {l n}^{2} 3 + 4 . l n 2 . (l n 3 + l n 2) =$ ${l n}^{2} 3 + 4 {l n}^{2} 2 + 4 . l n 2 . l n 3 = {(2 l n 2 + l n 3)}^{2}$ $,$ $x = \frac{- l n 3 \pm (2 l n 2 + l n 3)}{2 l n 2} = {\begin{cases} 1 \\ - 1 - \frac{l n 3}{l n 2} = - {l o g}_{2} 6 \end{cases}$
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