Invader12idn
Answered

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log4(x)+log4(x+15)=2

Sagot :

Konrad509idn
[tex]D:x>0 \wedge x+15>0\\ D:x>0 \wedge x>-15\\ D:x>0\\\\ \log_4x+\log_4(x+15)=2\\ \log_4x(x+15)=2\\ 4^2=x(x+15)\\ 16=x^2+15x\\ x^2+15x-16=0\\ x^2-x+16x-16=0\\ x(x-1)+16(x-1)=0\\ (x+16)(x-1)=0\\ x=-16 \vee x=1\\ -16\not \in D \Rightarrow \boxed{x=1}[/tex]
Albinaidn
Log4(x)+log4(x+15)=2
x>0;
x+15>0 => x>0
Log4(x)+log4(x+15)=2
Log4(x*(x+15))=Log4(16)x*(x+15)=16 
x1=1
x2=-16 
 х=1
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