Question

log ↓2(√x+1) / log ↓ 2 (3√x-41) = 3
нужно с полным решением
​

Answer 1

Ответ:

$x=49$

Объяснение:

$\displaystyle\\\frac{log_2(\sqrt{x}+1)}{log_2(\sqrt[3]{x-41}) }=3$

================================

ОДЗ:  

           $\begin{cases}\sqrt{x}+1 > 0\\\sqrt[3]{x-41} > 0\\x\geq 0\end{cases}\\\\\\\begin{cases}\sqrt{x} > -1\\x-41 > 0\\x\geq 0\end{cases}\\\\\\\begin{cases}x \in \mathbb{R} \ \ \ ( luboi\ x )\\x > 41\\x\geq 0\end{cases}$

                 $\Downarrow$

              $x > 41$

================================

$log_2(\sqrt{x}+1)=3\times log_2(\sqrt[3]{x-41})\\\\log_2(\sqrt{x}+1)= log_2\left((\sqrt[3]{x-41})^3\right)\\\\log_2(\sqrt{x}+1)= log_2(x-41})\\\\\sqrt{x}+1=x-41\\\\\sqrt{x}=x-41-1\\\\\sqrt{x}=x-42 \qquad | \ (\ )^2\\\\x=(x-42)^2\\\\x=x^2-84x+1764\\\\x^2-84x-x+1764=0\\\\x^2-85x+1764=0\\\\x^2-36x-49x+1764=0\\\\x(x-36)-49(x-36)=0\\\\(x-36)(x-49)=0\\\\x_1-36=0 \qquad x_2-49=0\\\\\boxed{x_1=36} \qquad \quad \boxed{x_2=49}$

ОДЗ:  $x > 41$. Тогда:

$\boxed{x=49}$