Log4(x)+log4(x+15)=2

2 answers:

7 0

$D:x>0 \wedge x+15>0\\
D:x>0 \wedge x>-15\\
D:x>0\\\\
\log_4x+\log_4(x+15)=2\\
\log_4x(x+15)=2\\
4^2=x(x+15)\\
16=x^2+15x\\
x^2+15x-16=0\\
x^2-x+16x-16=0\\
x(x-1)+16(x-1)=0\\
(x+16)(x-1)=0\\
x=-16 \vee x=1\\
-16\not \in D \Rightarrow \boxed{x=1}$

Airida [17]3 years ago

3 0

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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