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Question Number 33064 by artibunja last updated on 09/Apr/18

Commented by abdo imad last updated on 09/Apr/18

$$wemusthave2x+3>0and10x^2+13x+4>0$$$$\mathrm{\Delta }={13}^2-4\times 10\times 4=169-160=9>0$$$$x_1=\frac{-13+3}{20}=-\frac{1}{2},x_2=\frac{-13-3}{20}=\frac{-16}{20}=-\frac{4}{5}$$$$wemusthavex\in \left(\right]-\mathrm{\infty },-\frac{4}{5}[\cup ]-\frac{1}{2},+\mathrm{\infty }\left[\right)\cap ]-\frac{3}{2},+\mathrm{\infty }[$$$$\Rightarrow x\in ]-\frac{3}{2},-\frac{4}{5}[\cup ]-\frac{1}{2},+\mathrm{\infty }[$$$$\left(e\right)\iff \frac{ln(10x^2+13x+4)}{ln(2x+3)}=2\iff$$$$ln(10x^2+13x+4)=ln\left({(2x+3)}^2\right)\iff$$$$10x^2+13x+4=4x^2+12x+9\iff$$$$6x^2+x-5=0$$$$\mathrm{\Delta }=1-4.6.(-5)=121\Rightarrow x_1=\frac{-1+11}{12}=\frac{10}{12}=\frac{5}{6}$$$$x_2=\frac{-1-11}{12}=-1but-1\in D_{\left(e\right)},\frac{5}{6}\in D_{\left(e\right)}\Rightarrow$$$$-1and\frac{5}{6}arethesolutionsofthisequation.$$

Commented by abdo imad last updated on 09/Apr/18

$$forx=-1wehave2x+3=1so-1isnotsolution.$$

Answered by Rasheed.Sindhi last updated on 10/Apr/18

$${\log }_{2\mathrm{x}+3}({10\mathrm{x}}^2+13\mathrm{x}+4)=2$$$${(2\mathrm{x}+3)}^2={10\mathrm{x}}^2+13\mathrm{x}+4$$$${4\mathrm{x}}^2+12\mathrm{x}+9={10\mathrm{x}}^2+13\mathrm{x}+4$$$${6\mathrm{x}}^2+\mathrm{x}-5=0$$$${6\mathrm{x}}^2+6\mathrm{x}-5\mathrm{x}-5=0$$$$6\mathrm{x}(\mathrm{x}+1)-5(\mathrm{x}+1)=0$$$$(\mathrm{x}+1)(6\mathrm{x}-5)=0$$$$\mathrm{x}=-1\mid \mathrm{x}=\frac{5}{6}$$$$\mathrm{x}=-1\mathrm{makes}2\mathrm{x}+3=1\mathrm{which}\mathrm{may}\mathrm{not}$$$$\mathrm{be}\mathrm{the}\mathrm{base}\mathrm{of}\mathrm{logarithm}(\mathrm{as}\mathrm{sir}\mathrm{abdo}\mathrm{imad}$$$$\mathrm{wrote}\mathrm{in}\mathrm{his}\mathrm{solution})$$$$\therefore \mathrm{x}=\frac{5}{6}$$

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