Question

log4(x^2) + log2(x-5)=2

Answer 1

Свойство логарифма: $\log_{a^m} b^n = \frac{n}{m} \cdot \log_a b$

ОДЗ: $x^2 \ \textgreater \ 0, \ x-5\ \textgreater \ 0; \ \ \ \ x\ \textgreater \ 0, \ x\ \textgreater \ 5; \ \ \ \boxed{x\ \textgreater \ 5}$

 $\log_{2^2} x^2 +\log_2 (x-5)=\log_2 4; \ \ \ \frac{1}{2} \cdot \log_2 x^2 +\log_2 (x-5)=\log_2 4; \\ \\ \log_2 (x^2)^\frac{1}{2} +\log_2 (x-5)=\log_2 4; \ \ \ \log_2 x+\log_2 (x-5)=\log_2 4; \\ \\ \log_2 (x\cdot (x-5))=\log_2 4; \ \ \ x\cdot (x-5)=4; \ \ x^2 -5x =4; \\ \\ x^2-5x -4=0; \ \ x_{1,2} = \frac{5 \pm \sqrt{25 +16}}{2}=\frac{5 \pm \sqrt{41}}{2}; \\ \\ \boxed{x_1=\frac{5}{2} + \frac{\sqrt{41}}{2}} \ ; \ \ x_2=\frac{5}{2} - \frac{\sqrt{41}}{2}$
$x_2$не удовлетворяет ОДЗ