# Logaritmic equation

Logarithms problems.

### Logaritmic equation

Greetings.

I need your help solving this logaritmic equation.

I tried to equal the base of each log then convert it into one only log and i got to the result 2^(x^2-4)=x+3/x-3
thank you.
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### Re: Logaritmic equation

$$x_{1 }$$=1 ,$$x_{2 }$$=2

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### Re: Logaritmic equation

$$x^{2}$$$$log_{2 }$$$$\frac{3+x}{10}$$-$$x^{2}$$$$log_{\frac{1}{2} }$$(2+3x)=$$x^{2}$$-4+2$$log_{\sqrt{2} }$$$$\frac{3x^{2}+11x+6}{10}$$

Definition set
$$\begin{array}{|l} \frac{3+x}{10} >0 \\ 2+3x>0\\\frac{3x^{2}+11x+6}{10} >0\end{array}$$

$$\begin{array}{|l} x >-3 \\ x >-\frac{2}{3}\\x\in(-\infty;-3)\cup (-\frac{2}{3};+\infty)\end{array}$$ $$\Rightarrow$$ x$$\in$$(-$$\frac{2}{3}$$;+$$\infty$$)
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$$x^{2}$$$$log_{2 }$$$$\frac{3+x}{10}$$+$$x^{2}$$$$log_{2 }$$(2+3x)-2.2$$log_{2 }$$$$\frac{(3+x)(3x+2)}{10}$$=$$x^{2}$$-4

$$x^{2}$$[ $$log_{2 }$$$$\frac{3+x}{10}$$+$$log_{2 }$$(2+3x) ]-4$$log_{2 }$$$$\frac{(3+x)(3x+2)}{10}$$=$$x^{2}$$-4

$$x^{2}$$$$log_{2 }$$$$\frac{(3+x)(3x+2)}{10}$$-4$$log_{2 }$$$$\frac{(3+x)(3x+2)}{10}$$=$$x^{2}$$-4

($$x^{2}$$-4)$$log_{2 }$$$$\frac{(3+x)(3x+2)}{10}$$=$$x^{2}$$-4

1 case $$x_{1 }$$=-2 $$\notin$$ Def.set
2 case $$x_{2 }$$=2 $$\in$$ Def.set
3 case x$$\ne$$$$\pm$$2 $$\Rightarrow$$ $$log_{2 }$$$$\frac{(3+x)(3x+2)}{10}$$=1

$$\frac{9x+6+3x^{2}+2x}{10}$$=$$2^{1}$$ $$\Rightarrow$$ 3$$x^{2}$$+11x-14=0

D=121-4.3(-14) =289=$$17^{2}$$ ; $$x_{3,4 }$$=$$\frac{-11\pm17}{2.3}$$

$$x_{3 }$$=-$$\frac{28}{6}$$= -4,(6) $$\notin$$Def.set
$$x_{4 }$$=$$\frac{6}{6}$$=1 $$\in$$ Def.set
The answers $$x_{1 }$$ ,$$x_{3 }$$-are dropped .

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